University:
College Entrance Preparation
Course:
ACT Preparation (Science)
Academic year:

2024
Views:

426

Pages:

2
Author:

salcignofi11

Name: Hour: **Heat Calculations Worksheet #2** 1. How many kilojoules of heat are absorbed when 100.0 g of water is heated from 18°C to 85°C? 100 g * 4.184 J/g°C * (85-18)°C = 28,032 J = 28.03 kJ 2. A piece of stainless steel weighing 155 g absorbs 141 J of heat when its temperature is increased 17.8°C. What is the specific heat of steel? C = Q / (mΔT) = 141 J / (155 g * 17.8°C) = 0.511 J/g°C 3. How many joules of heat are given off when 50 g of water cools from 75°C to 25°C? Q = mcΔT = 50 g * 4.184 J/g°C * (25-75)°C = -10,460 J 4. How many calories are given off in Problem 3? 10460 J * 1 cal / 4.184 J = 2500 cal 5. Calculate the number of joules of heat energy needed to increase the temperature of 80.0 g of copper metal from 21.0°C to 80.0°C. The specific heat of copper is 0.382 J/g°C. Q = mcΔT = 80 g * 0.382 J/g°C * (80-21)°C = 1126.9 J 6. Calculate the specific heat of a substance given that 21.9 joules of heat are required to raise the temperature of 40.0 g of the substance from 95.6°C to 99.4°C. C = Q / (mΔT) = 21.9 J / (40 g * (99.4 - 95.6)°C) = 0.15 J/g°C 7. Calculate the specific heat of a substance given that 221 J of heat are required to raise the temperature of 55.0 g of the substance by 4.0°C. C = Q / (mΔT) = 221 J / (55 g * 4°C) = 1 J/g°C 8. Determine the heat absorbed by a 3.50 x 10^5 g block of concrete if its temperature rises from 35.2°C to 77.1°C. The specific heat capacity of concrete is 0.84 J/g°C. Q = mcΔT = 3.5 x 10^5 g * 0.84 J/g°C * (77.1 - 35.2)°C = 117,998,000 J Q = cmΔT 9. What is the temperature change if a 235 g piece of copper absorbs 35,000 joules of heat energy? (Specific heat of copper is 0.385 joule/g°C) ΔT = Q / (cm) = 35,000 J / (0.385 J/g°C * 235 g) = 386°C 10. Calculate the final temperature after 145 joules of heat energy is added to 7.73 g of water at 43.2°C (specific heat capacity of water = 4.184 J/g°C). Q = cm(Tf - Ti) 145 J = 4.184 J/g°C * 7.73 g * (Tf - 43.2°C) Tf = 47.78°C 11. Calculate the final temperature after 145 joules of heat energy is removed from 6.55 g of water at 34.2°C. Tf = Q / (cm) + Ti = -145 J / (4.184 J/g°C * 6.55 g) + 34.2°C = 28.9°C 12. Convert 153,000 joules to calories. 153,000 J * 1 cal / 4.184 J = 36,567 cal