Example 1:
How much heat energy is required to change 2 kg of ice at 0°C into water at 20°C? [Specific latent heat of fusion of water = 334 000 J/kg; specific heat capacity of water = 4200 J/(kg K).]
Answer:
m = 2kg
Specific latent heat of fusion of water, L = 334 000 J/kg
specific heat capacity of water = 4200 J/(kg K)
Energy needed to melt 2kg of ice,
Q1 = mL = (2)(334000) = 668000J
Energy needed to change the temperature from 0°C to °C.
Q2 = mcθ = (2)(4200)(20 - 0) = 168000J
Total energy needed = Q1 + Q2 = 668000 + 168000 = 836000J
How much heat energy is required to change 2 kg of ice at 0°C into water at 20°C? [Specific latent heat of fusion of water = 334 000 J/kg; specific heat capacity of water = 4200 J/(kg K).]
Answer:
m = 2kg
Specific latent heat of fusion of water, L = 334 000 J/kg
specific heat capacity of water = 4200 J/(kg K)
Energy needed to melt 2kg of ice,
Q1 = mL = (2)(334000) = 668000J
Energy needed to change the temperature from 0°C to °C.
Q2 = mcθ = (2)(4200)(20 - 0) = 168000J
Total energy needed = Q1 + Q2 = 668000 + 168000 = 836000J
Example 2:
Starting at 20°C, how much heat is required to heat 0.3 kg of aluminum to its melting point and then to convert it all to liquid? [Specific heat capacity of aluminium = 900J kg-1 °C-1; Specific latent heat of aluminium = 321,000 Jkg-1, Melting point of aluminium = 660°C]
Answer:
m = 0.3kg
Specific latent heat of fusion of aluminium, L = 321 000 J/kg
specific heat capacity of aluminium = 900 J/(kg K)
Energy needed to increase the temperature from 20°C to 660°C
Q1 = mcθ = (0.3)(900)(660 - 20) = 172,800J
Energy needed to melt 0.3kg of aluminium,
Q2 = mL = (0.3)(321000) = 96,300J
Total energy needed = Q1 + Q2 = 172,800 + 96,300 = 269,100J
Starting at 20°C, how much heat is required to heat 0.3 kg of aluminum to its melting point and then to convert it all to liquid? [Specific heat capacity of aluminium = 900J kg-1 °C-1; Specific latent heat of aluminium = 321,000 Jkg-1, Melting point of aluminium = 660°C]
Answer:
m = 0.3kg
Specific latent heat of fusion of aluminium, L = 321 000 J/kg
specific heat capacity of aluminium = 900 J/(kg K)
Energy needed to increase the temperature from 20°C to 660°C
Q1 = mcθ = (0.3)(900)(660 - 20) = 172,800J
Energy needed to melt 0.3kg of aluminium,
Q2 = mL = (0.3)(321000) = 96,300J
Total energy needed = Q1 + Q2 = 172,800 + 96,300 = 269,100J
Example 3:
How much heat must be removed by a refrigerator from 2 kg of water at 70 °C to convert it to ice cubes at -11°C? [Specific heat capacity of water = 4200J kg-1 °C-1; Specific latent heat of fusion of ice = 334,000 Jkg-1, specific heat capacity of ice = 2100 J/(kg K)]
Answer:
m = 2kg
Specific latent heat of fusion of water, L = 334,000 J/kg
Specific heat capacity of water, cw = 4,200 J/(kg K)
Specific heat capacity of ice, ci = 2,100 J/(kg K)
Energy to be removed to reduce the temperature from 70°C to 0°C (Freezing point of water)
Q1 = mcθ = (2)(4200)(70 - 0) = 588,000J
Energy needed to freeze 2kg of water,
Q2 = mL = (2)(334,000) = 668,000J
Energy to be removed to reduce the temperature from 0°C to -11°C
Q3 = mcθ = (2)(2100)(0 - (-11)) = 46,200J
Total energy needed = Q1 + Q2 + Q3 = 588,000 + 668,000 = 46,200J = 1,302,200J
How much heat must be removed by a refrigerator from 2 kg of water at 70 °C to convert it to ice cubes at -11°C? [Specific heat capacity of water = 4200J kg-1 °C-1; Specific latent heat of fusion of ice = 334,000 Jkg-1, specific heat capacity of ice = 2100 J/(kg K)]
Answer:
m = 2kg
Specific latent heat of fusion of water, L = 334,000 J/kg
Specific heat capacity of water, cw = 4,200 J/(kg K)
Specific heat capacity of ice, ci = 2,100 J/(kg K)
Energy to be removed to reduce the temperature from 70°C to 0°C (Freezing point of water)
Q1 = mcθ = (2)(4200)(70 - 0) = 588,000J
Energy needed to freeze 2kg of water,
Q2 = mL = (2)(334,000) = 668,000J
Energy to be removed to reduce the temperature from 0°C to -11°C
Q3 = mcθ = (2)(2100)(0 - (-11)) = 46,200J
Total energy needed = Q1 + Q2 + Q3 = 588,000 + 668,000 = 46,200J = 1,302,200J
