Call Now to Set Up Tutoring:

(888) 888-0446

GRE Subject Test: Chemistry : General Thermodynamics

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

Create An Account

All GRE Subject Test: Chemistry Resources

1 Diagnostic Test 152 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Heat And Temperature

Calculate the final temperature when a 44.0g $\displaystyle 44.0g$ sample of metal (specific heat of the metal= $0.52 \frac{J}{g^{o}C}$ $\displaystyle 0.52 \frac{J}{g^{o}C}$ ) at $45.0^{o}C$ $\displaystyle 45.0^{o}C$ is placed into 95.0g $\displaystyle 95.0g$ of water at $23.0^{o}C$ $\displaystyle 23.0^{o}C$ . (specific heat of water is $4.18\frac{J}{g^{o}C}$ $\displaystyle 4.18\frac{J}{g^{o}C}$ )

Possible Answers:

$132^{o}C$ $\displaystyle 132^{o}C$

$76^{o}C$ $\displaystyle 76^{o}C$

$17^{o}C$ $\displaystyle 17^{o}C$

$24^{o}C$ $\displaystyle 24^{o}C$

Correct answer:

$24^{o}C$ $\displaystyle 24^{o}C$

Explanation:

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1^oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$ $\displaystyle Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$ $\displaystyle C=\frac{Q}{m\times \bigtriangleup T}$

Heat is transferred from the metal to the water.

$-Q_{metal}=Q_{water}$ $\displaystyle -Q_{metal}=Q_{water}$

$-(mC\bigtriangleup T)=mC\bigtriangleup T$ $\displaystyle -(mC\bigtriangleup T)=mC\bigtriangleup T$

$-(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$ $\displaystyle -(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$

$-(44.0g\cdot 0.52\frac{J}{g^{o}C}\cdot (T_{f}-45.0^oC)=95.0g\cdot4.18\frac{J}{g^o}C(T_{f}-23^{o}C)$ $\displaystyle -(44.0g\cdot 0.52\frac{J}{g^{o}C}\cdot (T_{f}-45.0^oC)=95.0g\cdot4.18\frac{J}{g^o}C(T_{f}-23^{o}C)$

$-(22.9\frac{J}{^{o}C}(T_{f}-45.0^{o}C))=397\frac{J}{^{o}C}(T_{f}-23^{o}C)$ $\displaystyle -(22.9\frac{J}{^{o}C}(T_{f}-45.0^{o}C))=397\frac{J}{^{o}C}(T_{f}-23^{o}C)$

$-(22.9\frac{J}{^{o}C}\cdot T_{f}-22.9\frac{J}{^{o}C}\cdot 45.0^{o}C)=397\frac{J}{^{o}C}\cdot T_{f}-397\frac{J}{^{o}C}\cdot 23.0^{o}C$ $\displaystyle -(22.9\frac{J}{^{o}C}\cdot T_{f}-22.9\frac{J}{^{o}C}\cdot 45.0^{o}C)=397\frac{J}{^{o}C}\cdot T_{f}-397\frac{J}{^{o}C}\cdot 23.0^{o}C$

$-(22.9\frac{J}{^{o}C}\cdot T_{f}-1031J)=397\frac{J}{^{o}C}\cdot T_{f}-9131J$ $\displaystyle -(22.9\frac{J}{^{o}C}\cdot T_{f}-1031J)=397\frac{J}{^{o}C}\cdot T_{f}-9131J$

$-22.9\frac{J}{^{o}C}\cdot \ T_{f}+1031J=397\frac{J}{^{o}C}\cdot T_{f}-9131J$ $\displaystyle -22.9\frac{J}{^{o}C}\cdot \ T_{f}+1031J=397\frac{J}{^{o}C}\cdot T_{f}-9131J$

$-22.9\frac{J}{^{o}C}\cdot \ T_{f}-397\frac{J}{^{o}C}\cdot T_{f}=-1031J-9131J$ $\displaystyle -22.9\frac{J}{^{o}C}\cdot \ T_{f}-397\frac{J}{^{o}C}\cdot T_{f}=-1031J-9131J$

$-419\frac{J}{^{o}C}\cdot T_{f}=-10162J$ $\displaystyle -419\frac{J}{^{o}C}\cdot T_{f}=-10162J$

$T_{f}=\frac{-10162^{o}C}{-419.9}=24^{o}C$ $\displaystyle T_{f}=\frac{-10162^{o}C}{-419.9}=24^{o}C$

Report an Error

Example Question #1 : Heat And Temperature

Calculate the specific heat of 19.3 g $\displaystyle 19.3 g$ of metal that requires 353 J $\displaystyle 353 J$ of heat energy to raise its temperature from $17.0^{o}C$ $\displaystyle 17.0^{o}C$ to $44.0^{o}C$ $\displaystyle 44.0^{o}C$ ?

Possible Answers:

$0.677\frac{J}{g^{o}C}$ $\displaystyle 0.677\frac{J}{g^{o}C}$

$17\frac{J}{g^{o}C}$ $\displaystyle 17\frac{J}{g^{o}C}$

$0.581\frac{J}{g^{o}C}$ $\displaystyle 0.581\frac{J}{g^{o}C}$

$0.98\frac{J}{g^{o}C}$ $\displaystyle 0.98\frac{J}{g^{o}C}$

Correct answer:

$0.677\frac{J}{g^{o}C}$ $\displaystyle 0.677\frac{J}{g^{o}C}$

Explanation:

$C=\frac{Q}{m\times \bigtriangleup T}$ $\displaystyle C=\frac{Q}{m\times \bigtriangleup T}$

$C=\frac{Q}{m\bigtriangleup T}=\frac{353J}{(19.3g\times (44.0^{o}C-17.0^{o}C))}=0.677\frac{J}{g^{o}C}$ $\displaystyle C=\frac{Q}{m\bigtriangleup T}=\frac{353J}{(19.3g\times (44.0^{o}C-17.0^{o}C))}=0.677\frac{J}{g^{o}C}$

Report an Error

Example Question #31 : Thermodynamics And Phases

What was the final temperature of the water if a 33.0 g $\displaystyle 33.0 g$ sample of water absorbs 423 J $\displaystyle 423 J$ of heat energy and heats up from $23^{o}C$ $\displaystyle 23^{o}C$ ? (specific heat of water is $4.18 \frac{J}{g^{o}C}$ $\displaystyle 4.18 \frac{J}{g^{o}C}$ )

Possible Answers:

$71^{o}C$ $\displaystyle 71^{o}C$

$54^{o}C$ $\displaystyle 54^{o}C$

$28^{o}C$ $\displaystyle 28^{o}C$

$117^{o}C$ $\displaystyle 117^{o}C$

Correct answer:

$28^{o}C$ $\displaystyle 28^{o}C$

Explanation:

$C=\frac{Q}{m\times \bigtriangleup T}$ $\displaystyle C=\frac{Q}{m\times \bigtriangleup T}$

Rearranging gives,

$Q=mC\bigtriangleup T$ $\displaystyle Q=mC\bigtriangleup T$

$Q=mC(T_{f}-T_{i})$ $\displaystyle Q=mC(T_{f}-T_{i})$

$423J=19g\times4.18\frac{J}{g^{o}C}\times(T_{f}-23^{o}C)$ $\displaystyle 423J=19g\times4.18\frac{J}{g^{o}C}\times(T_{f}-23^{o}C)$

$423J=79\frac{J}{^{o}C}\times(T_{f}-23^{o}C)$ $\displaystyle 423J=79\frac{J}{^{o}C}\times(T_{f}-23^{o}C)$

$\frac{423J}{79\frac{J}{^{o}C}}=T_{f}-23^{o}C$ $\displaystyle \frac{423J}{79\frac{J}{^{o}C}}=T_{f}-23^{o}C$

$5.35^{o}C=T_{f}-23^{o}C$ $\displaystyle 5.35^{o}C=T_{f}-23^{o}C$

$T_{f}=28^{o}C$ $\displaystyle T_{f}=28^{o}C$

Report an Error

Example Question #41 : Physical Chemistry

A 20.5 g $\displaystyle 20.5 g$ metal at $37.1^{o}C$ $\displaystyle 37.1^{o}C$ was placed in 23.3 g $\displaystyle 23.3 g$ of water at $24.2^{o}C$ $\displaystyle 24.2^{o}C$ . The final temp of the water and metal was $26.0^{o}C$ $\displaystyle 26.0^{o}C$ . Assuming heat was not lost to the surroundings find the specific heat of the metal? The specific heat of water is $4.18\frac{J}{g^{o}C}$ $\displaystyle 4.18\frac{J}{g^{o}C}$ .

Possible Answers:

$0.897\frac{J}{g^{o}C}$ $\displaystyle 0.897\frac{J}{g^{o}C}$

$2.32\frac{J}{g^{o}C}$ $\displaystyle 2.32\frac{J}{g^{o}C}$

$1.2\frac{J}{g^{o}C}$ $\displaystyle 1.2\frac{J}{g^{o}C}$

$0.77\frac{J}{g^{o}C}$ $\displaystyle 0.77\frac{J}{g^{o}C}$

Correct answer:

$0.77\frac{J}{g^{o}C}$ $\displaystyle 0.77\frac{J}{g^{o}C}$

Explanation:

$C=\frac{Q}{m\times \bigtriangleup T}$ $\displaystyle C=\frac{Q}{m\times \bigtriangleup T}$

Heat is transferred from the metal to the water.

$-Q_{metal}=Q_{water}$ $\displaystyle -Q_{metal}=Q_{water}$

$-(mC\bigtriangleup T)=mC\bigtriangleup T$ $\displaystyle -(mC\bigtriangleup T)=mC\bigtriangleup T$

$-(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$ $\displaystyle -(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$

$-(20.5g\cdot C\cdot (26.0^{o}C-37.1^oC)=23.3g\cdot4.18\frac{J}{g^o}C(26.0^{o}C-24.2^{o}C)$ $\displaystyle -(20.5g\cdot C\cdot (26.0^{o}C-37.1^oC)=23.3g\cdot4.18\frac{J}{g^o}C(26.0^{o}C-24.2^{o}C)$

$-(-227.55g^{o}C\cdot C)=175.3092J$ $\displaystyle -(-227.55g^{o}C\cdot C)=175.3092J$

$227.55g^{o}C\cdot\ C=175.3092J$ $\displaystyle 227.55g^{o}C\cdot\ C=175.3092J$

$C=\frac{175.3092J}{227.55g^{o}C}=0.77\frac{J}{g^{o}C}$ $\displaystyle C=\frac{175.3092J}{227.55g^{o}C}=0.77\frac{J}{g^{o}C}$

Report an Error

Example Question #361 : Gre Subject Test: Chemistry

What was the final temperature of the water if a 112 g $\displaystyle 112 g$ sample of water absorbs 231 J $\displaystyle 231 J$ of heat energy and heats up from $25.0^{o}C$ $\displaystyle 25.0^{o}C$ ? (specific heat of water= $4.18\frac{J}{g^{o}C}$ $\displaystyle 4.18\frac{J}{g^{o}C}$ )

Possible Answers:

$134^{o}C$ $\displaystyle 134^{o}C$

$25.5^{o}C$ $\displaystyle 25.5^{o}C$

$43.2^{o}C$ $\displaystyle 43.2^{o}C$

$75^{o}C$ $\displaystyle 75^{o}C$

Correct answer:

$25.5^{o}C$ $\displaystyle 25.5^{o}C$

Explanation:

$C=\frac{Q}{m\times \bigtriangleup T}$ $\displaystyle C=\frac{Q}{m\times \bigtriangleup T}$

Rearranging gives,

$Q=mC\bigtriangleup T$ $\displaystyle Q=mC\bigtriangleup T$

$Q=mC(T_{f}-T_{i})$ $\displaystyle Q=mC(T_{f}-T_{i})$

$231J=112g\times4.18\frac{J}{g^{o}C}\times(T_{f}-25.0^{o}C)$ $\displaystyle 231J=112g\times4.18\frac{J}{g^{o}C}\times(T_{f}-25.0^{o}C)$

$231J=468.16\frac{J}{^{o}C}\times(T_{f}-25.0^{o}C)$ $\displaystyle 231J=468.16\frac{J}{^{o}C}\times(T_{f}-25.0^{o}C)$

$\frac{231J}{468.16\frac{J}{^{o}C}}=T_{f}-25.0^{o}C$ $\displaystyle \frac{231J}{468.16\frac{J}{^{o}C}}=T_{f}-25.0^{o}C$

$0.493^{o}C=T_{f}-25.0^{o}C$ $\displaystyle 0.493^{o}C=T_{f}-25.0^{o}C$

$T_{f}=25.5^{o}C$ $\displaystyle T_{f}=25.5^{o}C$

Report an Error

Example Question #1 : Heat And Temperature

A 10.2 g $\displaystyle 10.2 g$ metal at $72.9^{o}C$ $\displaystyle 72.9^{o}C$ was placed in 21.2 g $\displaystyle 21.2 g$ of water at $25.0^{o}C$ $\displaystyle 25.0^{o}C$ . The final temp of the water and metal was $53.3^{o}C$ $\displaystyle 53.3^{o}C$ . Assuming heat was not lost to the surroundings find the specific heat of the metal? The specific heat of water is $4.18\frac{J}{g^{o}C}$ $\displaystyle 4.18\frac{J}{g^{o}C}$ .

Possible Answers:

$465\frac{J}{g^{o}C}$ $\displaystyle 465\frac{J}{g^{o}C}$

$12.5\frac{J}{g^{o}C}$ $\displaystyle 12.5\frac{J}{g^{o}C}$

$2.6\frac{J}{g^{o}C}$ $\displaystyle 2.6\frac{J}{g^{o}C}$

$173\frac{J}{g^{o}C}$ $\displaystyle 173\frac{J}{g^{o}C}$

Correct answer:

$12.5\frac{J}{g^{o}C}$ $\displaystyle 12.5\frac{J}{g^{o}C}$

Explanation:

$C=\frac{Q}{m\times \bigtriangleup T}$ $\displaystyle C=\frac{Q}{m\times \bigtriangleup T}$

Heat is transferred from the metal to the water.

$-Q_{metal}=Q_{water}$ $\displaystyle -Q_{metal}=Q_{water}$

$-(mC\bigtriangleup T)=mC\bigtriangleup T$ $\displaystyle -(mC\bigtriangleup T)=mC\bigtriangleup T$

$-(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$ $\displaystyle -(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$

$-(10.2g\cdot C\cdot (53.3^{o}C-72.9^oC)=21.2g\cdot4.18\frac{J}{g^o}C(53.3^{o}C-25.0^{o}C)$ $\displaystyle -(10.2g\cdot C\cdot (53.3^{o}C-72.9^oC)=21.2g\cdot4.18\frac{J}{g^o}C(53.3^{o}C-25.0^{o}C)$

$-(-199.92g^{o}C\cdot C)=2507.83J$ $\displaystyle -(-199.92g^{o}C\cdot C)=2507.83J$

$199.92g^{o}C\cdot C=2507.83J$ $\displaystyle 199.92g^{o}C\cdot C=2507.83J$

$C=\frac{2507.83J}{199.92g^{o}C}=12.5\frac{J}{g^{o}C}$ $\displaystyle C=\frac{2507.83J}{199.92g^{o}C}=12.5\frac{J}{g^{o}C}$

Report an Error

Example Question #361 : Gre Subject Test: Chemistry

How much heat is absorbed when 150 g $\displaystyle 150 g$ of water goes from $25.0^{o}C$ $\displaystyle 25.0^{o}C$ to $35.0^{o}C$ $\displaystyle 35.0^{o}C$ ? (Specific heat of water= $4.18 \frac{J}{g^{o}C}$ $\displaystyle 4.18 \frac{J}{g^{o}C}$ )

Possible Answers:

$6270\ J$ $\displaystyle 6270\ J$

$2270\ J$ $\displaystyle 2270\ J$

$12\ J$ $\displaystyle 12\ J$

$70\ J$ $\displaystyle 70\ J$

Correct answer:

$6270\ J$ $\displaystyle 6270\ J$

Explanation:

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$C=\frac{Q}{m\times \bigtriangleup T}$ $\displaystyle C=\frac{Q}{m\times \bigtriangleup T}$

$Q=mC\bigtriangleup T$ $\displaystyle Q=mC\bigtriangleup T$

$Q=mC(T_{f}-T_{i})$ $\displaystyle Q=mC(T_{f}-T_{i})$

$Q=150g\cdot\ 4.18\frac{J}{g^{o}C}\cdot (35^{o}C-25^{o}C)$ $\displaystyle Q=150g\cdot\ 4.18\frac{J}{g^{o}C}\cdot (35^{o}C-25^{o}C)$

$Q=6270\ J$ $\displaystyle Q=6270\ J$

Report an Error

Example Question #11 : General Thermodynamics

Calculate the specific heat of 20.4 g $\displaystyle 20.4 g$ of metal that requires 410.4 J $\displaystyle 410.4 J$ of heat energy to raise its temperature from $20.46^{o}C$ $\displaystyle 20.46^{o}C$ to $64.4^{o}C$ $\displaystyle 64.4^{o}C$ ?

Possible Answers:

$0.21\frac{J}{g^{o}C}$ $\displaystyle 0.21\frac{J}{g^{o}C}$

$4.2\frac{J}{g^{o}C}$ $\displaystyle 4.2\frac{J}{g^{o}C}$

$1.6\frac{J}{g^{o}C}$ $\displaystyle 1.6\frac{J}{g^{o}C}$

$0.516\frac{J}{g^{o}C}$ $\displaystyle 0.516\frac{J}{g^{o}C}$

Correct answer:

$0.516\frac{J}{g^{o}C}$ $\displaystyle 0.516\frac{J}{g^{o}C}$

Explanation:

$C=\frac{Q}{m\times \bigtriangleup T}$ $\displaystyle C=\frac{Q}{m\times \bigtriangleup T}$

$C=\frac{Q}{m\bigtriangleup T}=\frac{410.4J}{(20.4g\times (64.4^{o}C-25.4^{o}C))}=0.516\frac{J}{g^{o}C}$ $\displaystyle C=\frac{Q}{m\bigtriangleup T}=\frac{410.4J}{(20.4g\times (64.4^{o}C-25.4^{o}C))}=0.516\frac{J}{g^{o}C}$

Report an Error

Example Question #42 : Physical Chemistry

Calculate the final temperature when a 14.7g $\displaystyle 14.7g$ sample of metal (specific heat of the metal= $0.72 \frac{J}{g^{o}C}$ $\displaystyle 0.72 \frac{J}{g^{o}C}$ ) at $37.0^{o}C$ $\displaystyle 37.0^{o}C$ is placed into 31.7g $\displaystyle 31.7g$ of water at $26.2^{o}C$ $\displaystyle 26.2^{o}C$ . (Specific heat of water is $4.18 \frac{J}{g^{o}C}$ $\displaystyle 4.18 \frac{J}{g^{o}C}$ )

Possible Answers:

36^oC $\displaystyle 36^oC$

18^oC $\displaystyle 18^oC$

54^oC $\displaystyle 54^oC$

27^oC $\displaystyle 27^oC$

Correct answer:

27^oC $\displaystyle 27^oC$

Explanation:

$C=\frac{Q}{m\times \bigtriangleup T}$ $\displaystyle C=\frac{Q}{m\times \bigtriangleup T}$

Heat is transferred from the metal to the water.

$-Q_{metal}=Q_{water}$ $\displaystyle -Q_{metal}=Q_{water}$

$-(mC\bigtriangleup T)=mC\bigtriangleup T$ $\displaystyle -(mC\bigtriangleup T)=mC\bigtriangleup T$

$-(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$ $\displaystyle -(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$

$-(14.7g\cdot 0.72\frac{J}{g^{o}C}\cdot (T_{f}-37.0^oC)=31.7g\cdot4.18\frac{J}{g^o}C(T_{f}-26.2^{o}C)$ $\displaystyle -(14.7g\cdot 0.72\frac{J}{g^{o}C}\cdot (T_{f}-37.0^oC)=31.7g\cdot4.18\frac{J}{g^o}C(T_{f}-26.2^{o}C)$

$-(10.584\frac{J}{^{o}C}(T_{f}-37.0^{o}C))=132.51\frac{J}{^{o}C}(T_{f}-26.2^{o}C)$ $\displaystyle -(10.584\frac{J}{^{o}C}(T_{f}-37.0^{o}C))=132.51\frac{J}{^{o}C}(T_{f}-26.2^{o}C)$

$-(10.584\frac{J}{^{o}C}\cdot T_{f}-10.584\frac{J}{^{o}C}\cdot 37.0^{o}C)=132.51\frac{J}{^{o}C}\cdot T_{f}-132.51\frac{J}{^{o}C}\cdot 26.2^{o}C$ $\displaystyle -(10.584\frac{J}{^{o}C}\cdot T_{f}-10.584\frac{J}{^{o}C}\cdot 37.0^{o}C)=132.51\frac{J}{^{o}C}\cdot T_{f}-132.51\frac{J}{^{o}C}\cdot 26.2^{o}C$

$-(10.584\frac{J}{^{o}C}\cdot T_{f}-391.61J)=132.51\frac{J}{^{o}C}\cdot T_{f}-3472J$ $\displaystyle -(10.584\frac{J}{^{o}C}\cdot T_{f}-391.61J)=132.51\frac{J}{^{o}C}\cdot T_{f}-3472J$

$-10.584\frac{J}{^{o}C}\cdot T_{f}+391.61J=132.51\frac{J}{^{o}C}\cdot T_{f}-3472J$ $\displaystyle -10.584\frac{J}{^{o}C}\cdot T_{f}+391.61J=132.51\frac{J}{^{o}C}\cdot T_{f}-3472J$

$-10.584\frac{J}{^{o}C}\cdot T_{f}-132.51\frac{J}{^{o}C}\cdot T_{f}=-3472J-391.61J$ $\displaystyle -10.584\frac{J}{^{o}C}\cdot T_{f}-132.51\frac{J}{^{o}C}\cdot T_{f}=-3472J-391.61J$

$-143.094\frac{J}{^{o}C}\cdot T_{f}=-3863.61J$ $\displaystyle -143.094\frac{J}{^{o}C}\cdot T_{f}=-3863.61J$

$T_{f}=\frac{-3863.61^{o}C}{-143.094}=27^{o}C$ $\displaystyle T_{f}=\frac{-3863.61^{o}C}{-143.094}=27^{o}C$

Report an Error

Example Question #43 : Physical Chemistry

How much heat is absorbed when 50 g $\displaystyle 50 g$ of water goes from $24.5^{o}C$ $\displaystyle 24.5^{o}C$ to $43.0^{o}C$ $\displaystyle 43.0^{o}C$ ? (Specific heat of water = $4.18 \frac{J}{g^{o}C}$ $\displaystyle 4.18 \frac{J}{g^{o}C}$ )

Possible Answers:

896J $\displaystyle 896J$

43J $\displaystyle 43J$

166J $\displaystyle 166J$

3867J $\displaystyle 3867J$

Correct answer:

3867J $\displaystyle 3867J$

Explanation:

$C=\frac{Q}{m\times \bigtriangleup T}$ $\displaystyle C=\frac{Q}{m\times \bigtriangleup T}$

Rearranging gives,

$Q=mC\bigtriangleup T$ $\displaystyle Q=mC\bigtriangleup T$

$Q=mC(T_{f}-T_{i})$ $\displaystyle Q=mC(T_{f}-T_{i})$

$Q=50g\cdot\ 4.18\frac{J}{g^{o}C}\cdot (43.0^{o}C-24.5^{o}C)$ $\displaystyle Q=50g\cdot\ 4.18\frac{J}{g^{o}C}\cdot (43.0^{o}C-24.5^{o}C)$

$Q=3867\ J$ $\displaystyle Q=3867\ J$

Report an Error

View Tutors

Erin
Certified Tutor

East Carolina University, Bachelor of Science, Neuroscience.

View Tutors

Kimberly
Certified Tutor

University of Connecticut, Bachelor of Chemistry, Chemistry. University of Connecticut, Doctor of Philosophy, Organic Chemist...

View Tutors

Yongmao
Certified Tutor

Nankai University, Bachelor of Science, Chemistry. Washington University in St Louis, Doctor of Philosophy, Organic Chemistry.

All GRE Subject Test: Chemistry Resources

1 Diagnostic Test 152 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

French Tutors in Miami, ISEE Tutors in San Francisco-Bay Area, French Tutors in San Francisco-Bay Area, English Tutors in San Francisco-Bay Area, MCAT Tutors in Boston, Algebra Tutors in Phoenix, Physics Tutors in San Diego, GMAT Tutors in San Diego, GMAT Tutors in Denver, SSAT Tutors in Philadelphia

Popular Courses & Classes

ISEE Courses & Classes in Seattle, Spanish Courses & Classes in Dallas Fort Worth, MCAT Courses & Classes in Los Angeles, GRE Courses & Classes in Miami, Spanish Courses & Classes in Washington DC, SSAT Courses & Classes in Phoenix, GMAT Courses & Classes in Phoenix, GRE Courses & Classes in Atlanta, ACT Courses & Classes in Philadelphia, GRE Courses & Classes in San Diego

Popular Test Prep

ISEE Test Prep in San Francisco-Bay Area, LSAT Test Prep in Houston, GMAT Test Prep in Chicago, GMAT Test Prep in New York City, ACT Test Prep in Philadelphia, LSAT Test Prep in Philadelphia, GRE Test Prep in Los Angeles, MCAT Test Prep in Phoenix, ISEE Test Prep in Philadelphia, MCAT Test Prep in Atlanta

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

GRE Subject Test: Chemistry : General Thermodynamics

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

All GRE Subject Test: Chemistry Resources

Example Questions

Example Question #1 : Heat And Temperature

Example Question #1 : Heat And Temperature

Example Question #31 : Thermodynamics And Phases

Example Question #41 : Physical Chemistry

Example Question #361 : Gre Subject Test: Chemistry

Example Question #1 : Heat And Temperature

Example Question #361 : Gre Subject Test: Chemistry

Example Question #11 : General Thermodynamics

Example Question #42 : Physical Chemistry

Example Question #43 : Physical Chemistry

All GRE Subject Test: Chemistry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: