Physics IC Mr. Tiesler Thermodynamics Practice Problems 1.) A

Name: __________________________
Date: ____________
Physics I C
Mr. Tiesler
Thermodynamics Practice Problems
1.) A certain mass of water was heated with 41,840 Joules, raising its temperature from 22.0 °C
to 28.5 °C. Find the mass of the water, in grams. (c of H2O = 4.18 J/g °C)
𝑄 = 41,840 𝐽
βˆ†π‘‡ = 28.5 ℃ βˆ’ 22.0 ℃ = 6.5 ℃
𝑐 = 4.18 𝐽⁄𝑔 βˆ™ ℃
𝑄 = π‘šπ‘βˆ†π‘‡
π‘š=
𝑄
41,840 𝐽
=
= 1,540 𝑔
π‘βˆ†π‘‡ (4.18 𝐽⁄𝑔 βˆ™ ℃)(6.5 ℃)
2.) What is the specific heat capacity of silver metal if 55.0 g of the metal absorbs 47.3 calories
of heat and the temperature rises 15.0°C? (1 cal = 4.186 J)
π‘š = 55.0 𝑔
4.186 𝐽
𝑄 = 47.3 π‘π‘Žπ‘™ (
) = 198 𝐽
1 π‘π‘Žπ‘™
βˆ†π‘‡ = 15.0 ℃
𝑐 =?
𝑄 = π‘šπ‘βˆ†π‘‡
𝑐=
𝑄
198 𝐽
=
= 0.240 𝐽⁄𝑔 βˆ™ ℃
π‘šβˆ†π‘‡ (55.0 𝑔)(15.0 ℃)
3.) A 7.5 kg bar of copper at 25.0 oC absorbs 3.5x104 J of heat. What is the final temperature of
the copper?
π‘š = 7.5 π‘˜π‘”
𝑇𝑖 = 25.0 ℃
𝑐 = 390 π½β„π‘˜π‘” βˆ™ ℃
𝑇𝑓 =?
𝑄 = 3.5π‘₯104 𝐽
𝑄 = π‘šπ‘βˆ†π‘‡
𝑇𝑓 =
𝑄
3.5π‘₯104 𝐽
+ 𝑇𝑖 =
+ 25.0 ℃ = 37.0 ℃
(7.5 π‘˜π‘”)(390 π½β„π‘˜π‘” βˆ™ ℃)
π‘šπ‘
Name: __________________________
Date: ____________
Physics I C
Mr. Tiesler
Thermodynamics Practice Problems
4.) How much heat energy is given out when 500.0 g of steam at 100.00C condense and cools to
50.00C?
𝑄 = ±π‘šπΏπ‘£ = βˆ’(0.500 π‘˜π‘”)(2260 π‘˜π½β„π‘˜π‘”) = βˆ’1130 π‘˜π½ = βˆ’1130π‘₯103 𝐽
𝑄 = π‘šπ‘βˆ†π‘‡ = (0.500 π‘˜π‘”)(4186 π½β„π‘˜π‘” βˆ™ ℃)(50.0 ℃ βˆ’ 100.0 ℃) = βˆ’1.04π‘₯105 𝐽
π‘„π‘‘π‘œπ‘‘π‘Žπ‘™ = βˆ’1130π‘₯103 𝐽 + (βˆ’1.04π‘₯105 𝐽) = βˆ’1.21π‘₯106 𝐽
5.) An automobile engine does 520 kJ of work and loses 220 kJ of energy as heat. What is the
change in the internal energy of the engine? Treat the engine, fuel, and exhaust gases as a closed
system.
π‘Š = βˆ’520 π‘˜π½
𝑄 = βˆ’220 π‘˜π½
π‘ˆ =?
βˆ†π‘ˆ = π‘Š + 𝑄 = βˆ’520 π‘˜π½ + (βˆ’220 π‘˜π½) = βˆ’740 π‘˜π½
6.) Calculate the work done on or by a system that absorbs 260 kJ of heat experiences a change
in internal energy of 157kJ.
𝑄 = 260 π‘˜π½
π‘ˆ = 157 π‘˜π½
π‘Š =?
βˆ†π‘ˆ = π‘Š + 𝑄
π‘Š = βˆ†π‘ˆ βˆ’ 𝑄 = 157 π‘˜π½ βˆ’ 260 π‘˜π½ = βˆ’103 π‘˜π½