Question

The specific heat of water is $4200\mathrm{J}{\mathrm{kg}}^{-1}{\mathrm{K}}^{-1}$ and the latent heat of ice is $3.4\times 105\mathrm{J}{\mathrm{kg}}^{-1}$.$100\mathrm{gm}$ of ice at $0°\mathrm{C}$ is placed in $200\mathrm{g}$ of water at $25°\mathrm{C}$. The amount of ice that will melt as the temperature of water reaches $0°\mathrm{C}$ is close to (in grams):

• A. $63.8$
• B. $64.6$
• C. $61.7$
• D. $69.3$

Answer 1

The correct option is C

$61.7$

The explanation for the correct option:

C: $61.7$

Heat loss by water when it cools down to $0^0\mathrm{C}$ is,

The formula used for the calculation:

$\begin{array}{rcl}\mathrm{Q}& =& \mathrm{mws}∆\mathrm{\theta }\\ & =& (200/1000)\times \left(4200\right)\times \left(25\right)\\ & =& 21000\mathrm{J}\end{array}$

Which is less than the amount of heat $\left({\mathrm{m}}_{\mathrm{i}}\mathrm{f}\right)$ required to melt ice completely.

$∆{\mathrm{m}}_{\mathrm{i}}\mathrm{L}=21000\mathrm{J}$

To find the amount of ice melt (∆m_i), take:

$\begin{array}{rcl}∆{\mathrm{m}}_{\mathrm{i}}& =& [21000/(3.4\times 105\left)\right]\mathrm{x}103\mathrm{gm}\\ & =& 61.7\mathrm{grams}\end{array}$

Therefore, option (C) $61.7$ is the correct option.