All High School Physics Resources
Example Questions
Example Question #4 : Understanding The Concept Of Energy Transfer
A piece of bread is burned on one side and uncooked on the other after being heated in a toaster. What was the most likely form of heat transfer to cause the burn?
Radiation
Induction
Conduction
We need to know more information to reach a conclusion
Convection
Conduction
Conduction is a form of heat transfer requiring direct contact between two objects. In the question, the burned side of the bread was likely in direct contact with the heat source, while the other was only in contact with hot air. The burned side is heated by conduction, while the uncooked side is heated by convection.
Convection is the transfer of heat through a fluid medium, namely a liquid or gas. Radiation is the transfer of heat through electromagnetic waves.
Example Question #5 : Understanding The Concept Of Energy Transfer
Which of the following methods of heat transfer requires the two objects to be touching?
Conduction
Radiation
Induction
Thermodynamic discharge
Convection
Conduction
Conduction is the form of heat transfer that requires direct contact between two objects.
Radiation is heat transfer via electromagnetic radiation. Convection uses a fluid medium, such as air or water, for heat transfer. Induction and thermodynamic discharge are not recognized types of heat transfer.
Example Question #1 : Understanding The Concept Of Energy Transfer
A marshmallow on a stick is placed above a fire, but not touching the fire. The marshmallow heats up and softens. How is the marshmallow being heated?
I. Conduction
II. Convection
III. Radiation
II and III
I and II
I, II, and III
III only
II only
II and III
Conduction occurs when heat is transferred by direct contact between two objects. Convection occurs when heat is transferred via contact between a fluid and an object. Radiation is heat transfer via electromagnetic radiation.
In this question, the fire is a source of heat and electromagnetic radiation. When the radiation from the fire impacts the marshmallow, it is being heated by radiation. The fire is also heating the surrounding air, creating fluid currents. The heated air is also transferring energy to the marshmallow, heating by convection.
There is no conduction in this example.
Example Question #3 : Heat
A Carnot Cycle operates using a heat reservoir at a temperature of and a cold reservoir at a temperature of . What is the operating efficiency, , of this cycle, to the nearest percent?
The efficiency of a Carnot Cycle is defined by:
In this calculation, temperatures are given in Kelvin. Since Celsius and Kelvin exist in a one-to-one ratio, a conversion is unnecessary unless desired. We will perform the conversion, since most classes will advise it.
Using these values, we can calculate the efficiency:
Convert this value to a percentage:
Example Question #11 : Thermodynamics
The temperature of an ideal gas is raised from to . If the volume remains constant, what was its initial pressure if the final pressure is ?
For this problem, use Gay-Lussac's law to set up a direct proportion between pressure and temperature. Note that this law only applies when volume is constant.
Plug in our given values and solve for the initial pressure.
Example Question #601 : High School Physics
of soup at cools down to after . If the specific heat of the soup is , how much energy does the soup release into the room?
The formula for heat energy is:
We are given the initial and final temperatures, mass, and specific heat. Using these values, we can find the heat released. Note that the time is irrelevant to this calculation.
That means that the soup "lost" of energy. This is the amount that it released into the room. The value is negative for the soup, the source of the heat, but positive for the room, which receives it.
Example Question #602 : High School Physics
of soup cools down to after . If the specific heat of the soup is , and it released of energy into the room, what was the initial temperature of the soup?
The formula for heat energy is:
We are given the final temperature, mass, specific heat, and heat released. Using these values, we can find the initial temperature. Note that the time is irrelevant to this calculation. Since heat is released from the soup, the net change in the soup's energy is negative. Since the soup is cooling, we expect our answer to be greater than .
Example Question #603 : High School Physics
of soup at cools down after . If the specific heat of the soup is , and it released of energy into the room, what is the final temperature of the soup?
The formula for heat energy is:
We are given the initial temperature, mass, specific heat, and heat released. Using these values, we can find the final temperature. Note that the time is irrelevant to this calculation. Since heat is released from the soup, the net change in the soup's energy is negative. Since the soup is cooling, we expect our answer to be less than .
Example Question #604 : High School Physics
An ice cube at melts. As it melts, constant temperature readings are taken and the sample maintains the temperature of throughout the melting process. Which statement best describes the energy of the system?
Energy is being used to convert the ice to water
Energy increases as the sample moves from a solid to a liquid
The form of water has no bearing on the system energy
We would need to know the mass of the ice cube to draw a conclusion
Energy of the system remains the same
Energy is being used to convert the ice to water
When an object changes phase, it requires energy called "latent heat." In this case, even though the temperature is remaining constant, the energy inside of the ice cube is decreasing as it expends energy to melt.
Example Question #605 : High School Physics
A silver spoon is placed in a cup of tea. If the spoon has a mass of and the tea has of mass, what is the final temperature of the spoon?
The equation for two items reaching a thermal equilibrium is given by describing a heat transfer. The heat removed from one object is equal to the heat added to the other.
We are given the specific heat values of each substance, as well as their masses. We also know the initial temperature of each substance. Use these terms in the equation to solve for the final temperature. Remember that the final temperature will be the same for each substance, since they will be in thermodynamic equilibrium.