100g of water at 25° C is poured into an insulated cup. 50g of ice at 0°C is added to the water. The water is stirred until the temperature of the water has fallen to 0°C. 18g of ice remains unmelted. The specific heat capacity of capacity of water is 4.2 J/g°C. Which value does this experiment give for the specific latent heat of fusion of ice? answer is 330J/g. but how?
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Answer:
Answer
Let the amount of ice is m gm.
According to the principal of calorimeter
heat taken by ice = heat given by water
∴20×2.1×m+(m−20)×334
=50×4.2×40
376m=8400+6680
m=40.1
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Answer:
Q lost water= mcΔt=100×42×25=10500J
Q gain ice=10500J
Q=ml
L=Q÷m
=10500÷(50-18)
=328.125
After rounding the anwser is 330J/g
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