Specific Heat Q = m c ΔT

Q Heat absorbed by a material Measured in Joules (J)

m Mass of the material Measured in grams (g)

c Specific heat, dependent upon the material Measure in J/g°C

ΔT Change in Temperature ΔT = Tfinal – Tinitial Measured in °C

The temperature of 150 grams of water is raised from 20°C to 25°C The temperature of 150 grams of water is raised from 20°C to 25°C. The specific heat of water is 4.186 J/g°C. How much heat did the water gain? Q Ra I C F

How much heat is needed to raise the temperature of 100 How much heat is needed to raise the temperature of 100.0 g of water by 85°C if the specific heat of water is 4.186 J/g°C? How much heat is absorbed by a 750-g iron skillet when its temperature is raised from 25°C to 125°C? The specific heat of iron is 0.449 J/g°C.

You just set up a new aquarium with 75,000 g of water and need to raise the temperature by 3.8°C, how much heat is needed? The specific heat of water is 4.18 J/g°C. Q Ra I C F Q m = 75,000 g c = 4.18 J/g°C ΔT = 3.8 °C Q = mc ΔT no Q = mcΔT = (75,000g)(4.18 J/g°C)(3.8°C) = 1,191,300 J

2. A jeweler is trying to release a diamond from its setting in a 10 2. A jeweler is trying to release a diamond from its setting in a 10.0 g silver ring. If the temperature has to change by 10.0°C, how much heat is needed? The specific heat of silver is 0.235 J/g°C. Q Ra I C F Q m = 10.0 g c = 0.235 J/g°C ΔT = 10.0 °C Q = mc ΔT no Q = mcΔT = (10.0 g)(0.235 J/g°C)(10.0°C) = 23.5 J

3. How much heat is needed to raise the temperature of 42 g of water by 3.0°C if the specific heat of water is 4.18 J/g°C? Q Ra I C F Q m = 42 g c = 4.18 J/g°C ΔT = 3.0 °C Q = mc ΔT no Q = mcΔT = (42 g)(4.18 J/g°C)(3.0°C) = 526.68 J

4. Gold has a specific heat of 0. 13 J/g°C 4. Gold has a specific heat of 0.13 J/g°C. If a sample of gold with a mass of 250 g undergoes a temperature increase of 4.0°C, how much heat does it absorb? Q Ra I C F Q m = 250 g c = 0.13 J/g°C ΔT = 4.0 °C Q = mc ΔT no Q = mcΔT = (250g)(0.13 J/g°C)(4.0°C) = 130 J

5. A piece of iron at a temperature of 145°C cools off to a temperature of 45.0°C. If the iron has a mass of 10.0 g and a specific heat of 0.449 J/g·°C, how much heat is given up? Q Ra I C F Q m = 10.0 g c = 0.449 J/g°C ΔT = 45°C - 145°C = -100°C Q = mc ΔT no Q = mcΔT = (10.0g)(0.449 J/g°C)(-100°C) = -449 J

6. The specific heat of copper is 0. 39 J/g·°C 6. The specific heat of copper is 0.39 J/g·°C. How much heat is needed to raise the temperature of 1000.0 g of copper from 25°C to 45.0°C? Q Ra I C F Q m = 1000 g c = 0.39 J/g°C ΔT = 45°C - 20°C = 20°C Q = mc ΔT no Q = mcΔT = (1000 g)(0.39 J/g°C)(20°C) = 7800 J

7. How much heat is absorbed by a 750 g iron skillet when its temperature rises from 25°C to 125°C? The specific heat of iron is 0.449 J/g°C. Q Ra I C F Q m = 750 g c = 0.449 J/g°C ΔT = 125°C - 25°C = 100°C Q = mc ΔT no Q = mcΔT = (750g)(0.449 J/g°C)(100°C) = 33,675 J

8. How much heat is needed to raise the temperature of 100 8. How much heat is needed to raise the temperature of 100.0 g of water by 85.0°C? The specific heat of water is 4.18 J/g°C. Q Ra I C F no Q = mcΔT = (100g)(4.18 J/g°C)(85°C) = 35,530 J Q m = 100 g c = 4.18 J/g°C ΔT = 85°C Q = mc ΔT

9. How much heat is given up when 450 g of water cool from 45°C to 20°C? The specific heat of water is 4.18 J/g°C. Q Ra I C F Q m = 450 g c = 4.18 J/g°C ΔT = 20°C - 45°C = -25°C Q = mc ΔT no Q = mcΔT = (450g)(4.18 J/g°C)(-25°C) = -47,025 J

Q = mcΔT What does each variable stand for? Activator Q = mcΔT What does each variable stand for?

You just set up a new aquarium with 45,000 g of water and need to raise the temperature by 2.5°C, how much heat is needed? The specific heat of water is 4.18 J/g°C. Q Ra I C F Q m = 45,000 g c = 4.18 J/g°C ΔT = 2.5 °C Q = mc ΔT no Q = mcΔT = (45,000g)(4.18 J/g°C)(2.5°C) = 470,250 J

2. A jeweler is trying to release a diamond from its setting in a 8 2. A jeweler is trying to release a diamond from its setting in a 8.0 g silver ring. If the temperature has to change by 12.5°C, how much heat is needed? The specific heat of silver is 0.235 J/g°C. Q Ra I C F Q m = 8.0 g c = 0.235 J/g°C ΔT = 12.5 °C Q = mc ΔT no Q = mcΔT = (8.0 g)(0.235 J/g°C)(12.5°C) = 23.5 J

3. How much heat is needed to raise the temperature of 38 g of water by 5.5°C if the specific heat of water is 4.18 J/g°C? Q Ra I C F Q m = 38g c = 4.18 J/g°C ΔT = 5.5 °C Q = mc ΔT no Q = mcΔT = (38g)(4.18 J/g°C)(5.5°C) = 873.62 J

4. Gold has a specific heat of 0. 13 J/g°C 4. Gold has a specific heat of 0.13 J/g°C. If a sample of gold with a mass of 350 g undergoes a temperature increase of 14.0°C, how much heat does it absorb? Q Ra I C F Q m = 350 g c = 0.13 J/g°C ΔT = 14.0 °C Q = mc ΔT no Q = mcΔT = (350g)(0.13 J/g°C)(14.0°C) = 637 J

5. A piece of iron at a temperature of 145°C cools off to a temperature of 45.0°C. If the iron has a mass of 10.0 g and a specific heat of 0.449 J/g·°C, how much heat is given up? Q Ra I C F Q m = 10.0 g c = 0.449 J/g°C ΔT = 45°C - 145°C = -100°C Q = mc ΔT no Q = mcΔT = (10.0g)(0.449 J/g°C)(-100°C) = -449 J

6. The specific heat of copper is 0. 39 J/g·°C 6. The specific heat of copper is 0.39 J/g·°C. How much heat is needed to raise the temperature of 1000.0 g of copper from 25°C to 45.0°C? Q Ra I C F Q m = 1000 g c = 0.39 J/g°C ΔT = 45°C - 20°C = 20°C Q = mc ΔT no Q = mcΔT = (1000 g)(0.39 J/g°C)(20°C) = 7800 J

7. How much heat is absorbed by a 750 g iron skillet when its temperature rises from 25°C to 125°C? The specific heat of iron is 0.449 J/g°C. Q Ra I C F Q m = 750 g c = 0.449 J/g°C ΔT = 125°C - 25°C = 100°C Q = mc ΔT no Q = mcΔT = (750g)(0.449 J/g°C)(100°C) = 33,675 J

8. How much heat is needed to raise the temperature of 100 8. How much heat is needed to raise the temperature of 100.0 g of water by 85.0°C? The specific heat of water is 4.18 J/g°C. Q Ra I C F no Q = mcΔT = (100g)(4.18 J/g°C)(85°C) = 35,530 J Q m = 100 g c = 4.18 J/g°C ΔT = 85°C Q = mc ΔT

9. How much heat is given up when 450 g of water cool from 45°C to 20°C? The specific heat of water is 4.18 J/g°C. Q Ra I C F Q m = 450 g c = 4.18 J/g°C ΔT = 20°C - 45°C = -25°C Q = mc ΔT no Q = mcΔT = (450g)(4.18 J/g°C)(-25°C) = -47,025 J