Question

A jar of tea is placed in sunlight until it reaches an equilibrium temperature of 30◦C . In an attempt to cool the liquid, which has a mass of 165 g , 136 g of ice at 0.0 ◦C is added. At the time at which the temperature of the tea is 23.5 ◦C , find the mass of the remaining ice in the jar. The specific heat of water is 4186 J/kg · ◦ C . Assume the specific heat capacity of the tea to be that of pure liquid water. Answer in units of g.

Answer 1

Given
Initial temperature of tea Ti = 30degrees
Mass of tea mt=165g=0.165kg
Mass of ice mi=136g=0.136kg
temperature of iceT' =0degrees
Specific heat of water cw =4186J/kg.0C
Let x be the mass of the ice that already melted.
The thermal energy of the tea at 30 C= the thermal energy of the tea + the water from the ice at 23.5 C plus the energy required to melt x.

"When ice melts, (the heat of fusion) as it would take to heat an equivalent mass of water by 80 °C, while its temperature remains a constant 0 °C."
4186 * 0.165 * 30 = 4186 * (0.165 + x) * 23.5 + 4186 * x * 80
0.165* 30 = 0.165 *23.5 + x * ( 23.5 + 80)
x = 0.165 * (30 - 23.5) / (23.5 + 80) = 0.01036 kg=10.36g
The remaining ice is 136- x = 125.64 g