Question

In order to cool 1 ton of water at 20°C in an insulated tank, a person pours 140 kg of ice at –5°C into the water. The specific heat of water at room temperature is c = 4.18 kJ/kg· °C, and the specific heat of ice at about 0°C is c = 2.11 kJ/kg°C. The melting temperature and the heat of fusion of ice at 1 atm are 0°C and 333.7 kJ/kg.

Answer 1

Answer:

$T = 6.31 ^0 C$

Explanation:

1 ton = 907.2 kg

so here heat given by water = heat absorbed by the ice

so by energy balance we will have

$Q_1 = Q_2$

$mL + ms\Delta T_1 + ms\Delta T_2 = ms\Delta T_3$

so we have

$140(2.11)(5) + 140(333.7) + 140(4.18)(T) = 907.2(4.18)(20 - T)$

$48195 + 585.2 T = 75841.9 - 3792.1 T$

$T = \frac{27646.92}{4377.3}$

$T = 6.31 ^0 C$