Answer to Question #162896 in Molecular Physics | Thermodynamics for harry

Question

A volume of 0.50 L of water at 16°C is put into an aluminum ice cube tray of mass 0.250 kg at the same temperature. How much energy must be removed from this system by the refrigerator to turn the water into ice at -8.0°C?

Answer 1

Solution:

Specific heat capacity of water is 4179.6 [$\frac{J}{(kg\cdot K)}$ ]

The enthalpy of fusion is 333.55 [$\frac{J}{g}$]

Specific heat of ice is 2.04 [$\frac{J}{(g\cdot K)}$]

0.5 liters = 0.5 kg

0.5 kg * 4179.6 $\frac{J}{(kg\cdot K)}$ * (16 - 0) K + 0.5 kg * 333.55 $\frac{J}{g}$   +  0.5 kg * 2.04 $\frac{J}{(g\cdot K)}$ * (0 - (-8)) K =>

0.5 * 4179.6 * 16 J + 0.5 * 333.55 kJ + 0.5 * 2.04 * 8 kJ =>

0.5 * (4.1796 * 16 + 333.55 + 2.04 * 8) kJ =>

0.5 * 416.7436 kJ => 208.3718 kJ

Specific heat of Aluminum is 0.9 $(\frac{J}{g\cdot K})$

E_w = 0.500*(4.186*16+333+2.09*8) = 208.3 kjoule

E_a = 0.25*0.9*(16+8) = 5.3 kjoule

E = E_w+E_a = 208.3 + 5.3 = 213.6 kjoule =214 kJ

Answer:

214 kJ