Question

Ice at $-20{}^{\circ }\text{C}$ is added to $50\text{g}$ of water at $40{}^{\circ }\text{C}$, when the temperature of the mixture reaches $0{}^{\circ }\text{C}$, it is found that $20\text{g}$ of ice is still unmelted. The amount of ice added to the water was close to
$\text{(Specific heat of water}=4.2{\text{Jg}}^{-1}{}^{\circ }{\text{C}}^{-1}\text{Specific heat of Ice}=2.1{\text{Jg}}^{-1}{}^{\circ }{\text{C}}^{-1}\text{Heat of fusion of water at}0{}^{\circ }\text{C}=334{\text{Jg}}^{-1})$

• A. $50\text{g}$
• B. $100\text{g}$
• C. $60\text{g}$
• D. $40\text{g}$

Answer 1

The correct option is D $40\text{g}$

Let $m$ gram of ice is added.

From principal of calorimeter

$\text{heat gained by ice = heat lost by water}$

$\therefore m{s}_i\Delta T_{ice}+(m-20)L_f=m_w{S}_w\Delta T_{water}$

$\Rightarrow m\times 2.1\times 20+(m-20)\times 334=50\times 4.2\times 40$

$\Rightarrow 376m=8400+6680$

$\Rightarrow m\approx 40\text{g}$

Hence, option $\left(D\right)$ is correct.