Temperature and Heat

Assessment Problems

• 1. A sample of water at 25o C is heated until the temperature is doubled to 50o C. Will the Fahrenheit temperature double as well? Explain your answer.

Answer: The Fahrenheit temperature will not be doubled. The freezing point of water on the Celsius scale is 00 while the boiling point of water on that scale is 1000. On the Fahrenheit scale, the values are 320 and 2120 respectively. Since the difference between the two points on the Fahrenheit scale is 180 and the difference on the Celsius scale is 100, there are 180/100 or 1.8 Fahrenheit degree for every Celsius degree. Since they aren’t in a one to one correspondence, doubling the Celsius temperature won’t result in a doubling of the Fahrenheit temperature.

• 2. 500 g of water at 25oC is heated by absorbing 1200 calories of heat.  Calculate the final temperature of the water sample.

Answer: Q = mcw(tf – ti)

1200 cal = (500 g)(1 cal/g-C0)(tf – 250 C)

1200 = 500( tf – 25)

1200 = 500tf – 12500

500 tf = 13700

tf = 27.40 C

• 3. Explain why you feel warm sitting in a car on a cold winter day.

Answer: If the windows are closed, the inside of the car will warm up due to the “greenhouse effect”. The sun passing through the glass contains energy. The seats and other types of material inside the car absorb this energy which then heats the air inside the car. This can cause potential harm in the summer time when the sun is higher in the sky, especially to pets left inside the car. In the wintertime, the sun is not overhead and less energy enters the car.

• 4. A 20 g sample of water and a 20 g sample of copper each absorb 1000 calories of thermal energy. Which sample undergoes the largest increase in temperature? The specific heat of water is 1 cal/g-0C and that of copper is 0.092 cal/g-0C.

Answer: The sample of copper undergoes the largest increase in temperature. This is because it has a smaller specific heat. Less energy is required to raise the temperature of 1 gram of copper 1 degree Celsius.

For copper, 1000 cal = (0.092 cal/g-0C)(change in temp.)(20 g).  Solving for the change in temperature, you should calculate 5400 C.

For water, 1000 cal = (1 cal/g-0C)(change in temp.)(20 g).  In this case, the change in temperature will be only 500 C.

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