In a closed system, energy and work can be exchanged with the surrounding area but matter is contained. Consider, for example, a beaker of ice. Heat from the surroundings it transferred to the ice, causing it to melt, but there is no exchange of matter. No external compounds react with the ice and none of the ice leaves the beaker.

In an open system, both heat and matter can be exchanged. In an isolated system neither heat, nor matter is exchanged.

This process can be divided into three stages: the raising of the temperature of the ice, the melting of the ice into water, and the raising of the temperature of the water. The first and third step use the specific heat of the substance to calculate the heat required, while the second step uses the heat of fusion.

Step 1 and 3: \(\displaystyle q=mc\Delta T\)

Step 2: \(\displaystyle q=m\Delta H_{fus}\)

We know the mass of the water, the change in temperature, and the constant values. Using these values, we can sum the heat required for each step to get our final answer.

\(\displaystyle q_{total}=q_1+q_2+q_3\)

\(\displaystyle q_1=(12.0g)(2.03\frac{J}{g^oC})(0^0C-(-3^oC))=73.08J\)

\(\displaystyle q_2=(12.0g)(334\frac{J}{g})=4008J\)

\(\displaystyle q_3=(12.0g)(4.18\frac{J}{g^oC})(15^oC-0^oC)=752.4J\)

\(\displaystyle q_{total}=73.08J+4008J+752.4J=4833.48J\)

There are two processes requiring added heat in this problem:

1. Raising the temperature of the liquid water from \(\displaystyle \small 20^oC\) to \(\displaystyle \small 100^oC\) (use \(\displaystyle \Delta Q = mc\Delta T\))

2. Boiling the water at a constant temperature of \(\displaystyle \small 100^oC\) (use \(\displaystyle \Delta Q = m\Delta H^o_{vap}\))

To use either of these equation, we need to find the mass of the water using the relation between mass, density, and volume.

\(\displaystyle m = \rho*V= 1\frac{g}{mL}*1000mL=1000g\)

Use this mass with the given specific heat and temperatures to find the heat for part 1 of the process.

\(\displaystyle \Delta Q = (1000g)(4.186\frac{J}{g^oC})(100^oC - 20^oC)=334,880J\)

Then, use the mass with the given heat of vaporization to find the energy needed to convert the water to water vapor.

\(\displaystyle \Delta Q =(1000g) (2260\frac{J}{g})= 2,260,000J\)

Sum the energies for step 1 and step 2.

\(\displaystyle 334,880 J + 2,260,000 J = 2,594,880J\approx 2,595kJ\)

This is the total amount of energy needed from the burning wood. Use stoichiometry to find the grams of wood needed to produce this amount of energy.

\(\displaystyle 2595kJ*(\frac{1 g}{20kJ})=129.7g\approx 130g\)

If heat is transferred from the metal object to the water, that means that the temperature of the metal must be greater than that of the water. We must find the answer choice where this is the case. Since the temperatures are given in different units, they must be converted to the same unit for comparison purposes.

\(\displaystyle K=\ ^oC+273\)

\(\displaystyle ^oF=\frac{9}{5}(^oC)+32\)

Only one answer option results in the metal having a greater temperature than the water:

\(\displaystyle \text{Water: }65^oC\ \ \text{Metal: }340K\)

\(\displaystyle 65^oC+273=338K\rightarrow T_{water}< T_{metal}\)

In this problem, we need to track the transfer of heat from the aluminum to the water. Since the heat acquired by the water is equal to the heat given off by the aluminum, we can set their equations equal to each other; however, in order to avoid a negative number, aluminum's change in temperature will be set as the initial temperature minus the final temperature. This results in the equation below.

\(\displaystyle q=mc\Delta T\)

\(\displaystyle m_{Al}c_{Al}(T_{initial}-T_{final}) = m_{H_2O}c_{H_2O}(T_{final}-T_{initial})\) \(\displaystyle \small = q\)

\(\displaystyle \large \small (35 g)(0.9\frac{J}{g*K})(373K-T_{Final}) = (150 g)(4.184\frac{J}{g*K})(T_{Final}-298K)\)

\(\displaystyle \small T_{final } = 301.6K\)

When a specific heat capacity is given, we typically use the equation \(\displaystyle q = mc\Delta T\). Since we know all of the factors except for the change in temperature, we can simply solve for \(\displaystyle \Delta T\).

\(\displaystyle 1200J = (200g)(0.13\frac{J}{g*K})\Delta T\)

\(\displaystyle \Delta T=\frac{1200J}{(200g)(0.13\frac{J}{g*K})}\)

\(\displaystyle \Delta T = 46.15K\)

The equation for enthalpy of reaction is:

\(\displaystyle \Delta H_{reaction}^{0} = \Delta H_{f, products}^{0} - \Delta H_{f, reactants}^{0}\)

Given our chemical reaction and the enthalpies of formation, we can find the enthalpy of reaction.

\(\displaystyle 2C_{8}H_{18}+25O_{2}\rightarrow 16CO_{2} +18H_{2}O\)

First, find the total enthalpy for the products.

\(\displaystyle \Delta H_{f, products}^{0} = (16 ) \Delta H_{f, CO_{2}} + (18 ) \Delta H_{f, H_{2}O}\)

\(\displaystyle \Delta H_{f, products}^{0} = (16 ) (-394kJ) + (18 )(-286kJ) = -11452 kJ\)

Then, find the total enthalpy for the reactants.

\(\displaystyle \Delta H_{f, reactants}^{0} = (2) \Delta H_{f, C_{8}H_{18}} + (25 ) \Delta H_{f, O_{2}}\)

\(\displaystyle \Delta H_{f, reactants}^{0} =( 2) (269kJ) + (25) (0kJ) = 538kJ\)

Since the oxygen is elemental, its heat of formation is zero.

Return to the original equation to calculate the final enthalpy of reaction.

\(\displaystyle \Delta H_{reaction}^{0}=-11452kJ-(538kJ)=-11990kJ\approx-1.2*10^{4}kJ\)

Hess's law states that the enthalpies of the steps in an overall reaction can be added in order to find the overall enthalpy. Since the overall enthalpy of the reaction is +68kJ, we can use the following equation to find step 2's enthalpy.

\(\displaystyle Step 1 \Delta H + step 2 \Delta H = overall reaction \Delta H\)

\(\displaystyle 180kJ + step 2 \Delta H = +68kJ\)

\(\displaystyle Step 2 \Delta H = -112kJ.\)

1)  \(\displaystyle A \rightarrow 2B + C + \Delta H_{1}\)

2)  \(\displaystyle D\rightarrow B + C + \Delta H_{2}\)

3)  \(\displaystyle D \rightarrow 2E + \Delta H_{3}\)

\(\displaystyle A + C \rightarrow 4E\)

We need to add the given processes in such a way that leaves \(\displaystyle \small A+C\) on the left side, and \(\displaystyle \small 4E\) on the right. 

Since there are no \(\displaystyle \small B's\) in the final equation, begin by combining processes 1 and 2 to eliminate.

First, reverse process 2 and multiply by two:

2a) \(\displaystyle 2 (B + C + \Delta H_{2} \rightarrow D)\)

2a) \(\displaystyle 2B + 2C + 2\Delta H_{2} \rightarrow 2 D\)

2a) \(\displaystyle 2B + 2C \rightarrow 2 D - 2\Delta H_{2}\)

Add this result to process 1.

1) \(\displaystyle A \rightarrow 2B + C +\Delta H_{1}\)

2a) \(\displaystyle 2B + 2C \rightarrow 2D -2\Delta H_{2}\)

1+2a) \(\displaystyle A + C \rightarrow 2D + \Delta H_{1} -2\Delta H_{2}\)

Now we need to combine with process 3 to eliminate the \(\displaystyle \small D's\). Start by multiplying process 3 by two.

3a) \(\displaystyle 2(D \rightarrow 2E + \Delta H_{3})\)

3a) \(\displaystyle 2D \rightarrow 4E + 2\Delta H_{3}\)

Add this to the combined result from processes 1 and 2.

1+2a) \(\displaystyle A + C \rightarrow 2D + \Delta H_{1} -2\Delta H_{2}\)

3a) \(\displaystyle 2D \rightarrow 4E + 2\Delta H_{3}\)

1+2a+3a) \(\displaystyle A + C \rightarrow 4E +\Delta H_{1} - 2\Delta H_{2} + 2\Delta H_{3}\)

This gives our final combination. The elemental reaction is \(\displaystyle A + C \rightarrow 4E\), and the enthalpy of the process is \(\displaystyle \small \Delta H_1-2\Delta H_2+2\Delta H_3\).

Scenarios 3 and 4 are presented as thought experiments because, common sense tells us, they will never happen. The fundamental reason they will never happen is because they do not lead to a total increase in entropy of the universe. In order for an event to proceed, this requirement must be satisfied.