10 g of ice cubes at 0⁰C are released in a tumbler (water equivalent 55 g) at 40⁰C. Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearly (Take, L = 80 cal/g)

Let T be the temperature when thermal equilibrium has reached.

Heat gained by ice to be converted to water at 0⁰C

$=mL+m\times c\times (T-0)=10\times 80+10\times 1\times T$

Heat lost by tumbler and its contents = 55 x (40 $-$ T)

Using principle of calorimetry, heat gained = heat lost

$10\times 80+10\times 1\times T=55\times (40-T)$

$\Rightarrow 65T=2200-800=1400$

$\Rightarrow T=\frac{1400}{65}\approx {21.5}^{\circ }\mathrm{C}\approx {22}^{\circ }\mathrm{C}$