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Thermal Properties of Matter

JEE Main level, single correct, medium

75 of 126

Question 75 of 126

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

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Answer: (D) $\dfrac{5{0M}_{2}}{{M}_{1}}-5$

Principle of calorimetry

{M}_{1}{S}_{ice}\left(0-\left(-10\right)\right)+{M}_{1}{L}_{f}={M}_{2}{S}_{water}\left(50-0\right)

{M}_{1}\left(0.5\right)\left(10\right)+{M}_{1}{L}_{f}={M}_{2}\left(50\right)

5{M}_{1}+{M}_{1}{L}_{f}={M}_{2}\left(50\right)

{L}_{f}=\dfrac{50{M}_{2}}{{M}_{1}}-5

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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