Question #232920
How much heat is necessary to change 500g of ice at 263K to water at 293K
1
Expert's answer
2021-09-03T13:56:29-0400

m=500  gT1=263273=10  °CT2=293273=20  °Cm=500 \;g \\ T_1= 263-273 = -10 \;°C \\ T_2 = 293 - 273 = 20 \; °C

Specific heat capacity for Ice = 2.1 J/g°C

Latent heat of fusion of Ice = 335 J/g

Specific heat capacity for Water = 4.2 J/g°C

Q1=10×500×2.1=10500  JQ_1 = 10\times 500 \times 2.1 = 10500 \;J (from -10 °C to 0 °C)

Q2=20×500×4.2=42000  JQ_2 = 20 \times 500 \times 4.2 = 42000 \;J (from 0 °C to 20 °C)

Q3=500×335=167500  JQ_3 = 500 \times 335 = 167500 \;J (heat of melting)

Total heat QT=10500+42000+167500=220000  JQ_T = 10500+42000+167500 = 220000\;J

Answer: 220 kJ


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