An ideal refrigerator is working between the temperature of ice and temperature of atmosphere at 300 K. Find the energy which has been supplied to it to freeze 2 kg of water at 0°C. Given that latent heat of ice 3.33 × 105 s J/kg.
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4
Answer:
Total heat = Heat required to convert 2 kg of ice to 2 kg of water at 0 °C + Heat required to convert 2 kg of water at 0 °C to 2 kg of water at 20 °C.
Heat=mhfg+mCpΔT
Answered by
7
Answer:
Therefore, to melt 2kg of ice 835.48kJ of heat is required.
Explanation:
Total heat = Heat required to convert 2kg of ice to 2kg of water at 0℃ +Heat required to convert 2kg of water at 0℃ to 2kg of water at 20℃
Heat=mhfg+mCp\Delta t$$
Here, m(mass of ice)=2kg
hfg(latent heat of fusion of ice)=334kJ
Cp of water(specific heat)=4.187 kJ/kg-k
Δt(temperature difference) =20℃
Therefore,
Heat required=2×334+2×4.187×(20−0)
=835.48 kJ
Therefore, to melt 2kg of ice 835.48kJ of heat is required.
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