Specific heat of water =4200, J, kg ^{-1}, K ^{-1} and latent heat of ice =3.4 × 10^{5}, J, kg ^{-1

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10-1.Thermometry, Thermal Expansion and Calorimetry

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The specific heat of water $=4200\, J\, kg ^{-1}\, K ^{-1}$ and the latent heat of ice $=3.4 \times 10^{5}\, J\, kg ^{-1}.$ $100$ grams of ice at $0^{\circ} C$ is placed in $200\, g$ of water at $25^{\circ} C$. The amount of ice that will melt as the temperature of water reaches $0^{\circ} C$ is close to (in $grams$)

A

$61.7$

B

$63.8$

C

$69.3$

D

$64.6$

(JEE MAIN-2020)

Solution

Here the water will provide heat for ice to melt therefore

$m_{w} s_{w} \Delta \theta=m_{i c e} L_{i c e}$

$m _{ ice }=\frac{0.2 \times 4200 \times 25}{3.4 \times 10^{5}}$

$=0.0617 \,kg$

$=61.7\, gm$

Remaining ice will remain un-melted

Standard 11

Physics

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