When $\mathrm{M}_{1}$ gram of ice at $-10^{\circ} \mathrm{C}$ (specific heat $=0.5 \mathrm{cal} \mathrm{g}^{-1 \circ} \mathrm{C}^{-1}$ ) is added to $\mathrm{M}_{2}$ gram of water at $50^{\circ} \mathrm{C}$, finally no ice is left and the water is at $0^{\circ} \mathrm{C}$. The value of latent heat of ice, in cal $\mathrm{g}^{-1}$ is:

(A) $\frac{50 \mathrm{M}_{2}}{\mathrm{M}_{1}}-5$

(B) $\frac{5 \mathrm{M}_{2}}{\mathrm{M}_{1}}-5$

(D) $\frac{5 \mathrm{M}_{1}}{\mathrm{M}_{2}}-50$

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Heat & Thermodynamics

When

M_{1}

gram of ice at

- 1 0^{\circ} C

(specific heat

= 0.5 cal g^{- 1 \circ} C^{- 1}

) is added to

M_{2}

gram of water at

5 0^{\circ} C

, finally no ice is left and the water is at

0^{\circ} C

. The value of latent heat of ice, in cal

g^{- 1}

is:

Answer: (A)

Use calorimetry: the heat the warm water gives up equals the heat the ice takes in. Cooling

M_{2}

grams of water from

5 0^{\circ} C

0^{\circ} C

releases

M_{2} \times 1 \times 50 = 50 M_{2}

cal. The ice first warms from

- 1 0^{\circ} C

0^{\circ} C

, absorbing

M_{1} \times 0.5 \times 10 = 5 M_{1}

cal, and then melts, absorbing

M_{1} L

cal. Balancing,

50 M_{2} = 5 M_{1} + M_{1} L

, so

M_{1} L = 50 M_{2} - 5 M_{1}

and the latent heat is

L = \frac{50 M _{2}}{M _{1}} - 5 cal g^{- 1}

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option