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When 50 gram of ice at 0 0C is mixed with 300 gram of water at 50 0C, the resulting temperature is 6.7 0C Calculate the heat of fusion of ice?

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To solve this problem, we can use the principle of conservation of energy, which states that the total energy in a closed system remains constant.
First, we need to calculate the total heat gained or lost by each substance in the system. The heat gained or lost can be calculated using the formula:
Q = mcΔT
where:
Q = heat gained or lost
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature
For the ice:
m = 50 g
c = 2.09 J/g°C (specific heat capacity of ice)
ΔT = 6.7°C - 0°C = 6.7°C
Q_ice = 50 g * 2.09 J/g°C * 6.7°C = 706.15 J
For the water:
m = 300 g
c = 4.18 J/g°C (specific heat capacity of water)
ΔT = 6.7°C - 50°C = -43.3°C (the water is losing heat)
Q_water = 300 g * 4.18 J/g°C * -43.3°C = -5630.94 J
Since the total energy in a closed system remains constant, the heat lost by the water must be equal to the heat gained by the ice:
Q_ice = -Q_water
706.15 J = 5630.94 J
5630.94 J - 706.15 J = 4934.79 J
This heat corresponds to the heat of fusion of ice. Therefore, the heat of fusion of ice is 4934.79 J.
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