A 73.8-g aluminum ice tray in a home refrigerator holds 323 g of water. Calculat

Question

A 73.8-g aluminum ice tray in a home refrigerator holds 323 g of water. Calculate the energy in kJ that must be removed from the tray and its contents to reduce the temperature from 19.6°C to 0.0°C, freeze the water, and drop the temperature of the tray and ice to –11.0°C. Assume the specific heat of aluminum remains constant at 0.900 J/g·°C over the temperature range involved. Data for water at 1 atm: Melting point = 0.0°C Specific heat liquid = 4.18 J/g·°C Specific heat solid = 2.06 J/g·°C Heat of fusion = 333 J/g

__kJ

Explanation / Answer

Qaluminium = m*C*(Tf-Ti)

Qaluminium = 73.8*0.9 * (-11 - 19.6) = -2032.452 J for Al

now, for water

Qwater (to 0) = m*C*(Tf-ti) = (323*4.184)(0-19.6)

Qfreezing = m*LH = 323*-333 =

Qice = m*C*(Tf-Ti) = 323*2.02*(-11-0)

Qwater = (323*4.184)(0-19.6) +  323*-333 + 323*2.02*(-11-0) = -141224.1272 J

Qtotal --> Qwater + Qal = -141224.1272 +-2032.452

Qtotal = -143256.57 J

Q(kJ) = -143.25 kJ

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