Thermal Properties of Matter

JEE Main level, single correct, medium

58 of 126

Question 58 of 126

Heat energy of 184kJ184kJ is given to ice of mass 600g600g at −12∘C-12^{\circ}C, Specific heat of ice is 2222.3J{kg}^{–1}^{\circ}{C}^{–1} and latent heat of ice is 336kJkg–1336kJ{kg}^{–1}.
(A) Final temperature of system will be 0∘C0^{\circ}C
(B) Final temperature of the system will be greater than 0∘C0^{\circ}C
(C) The final system will have a mixture of ice and water in the ratio of 5:15:1
(D) The final system will have a mixture of ice and water in the ratio of 1:51:5
(E) The final system will have water only
Choose the correct answer from the options given below :

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Answer: (A) A and D only

Given:
∆Q=184×103J∆Q=184\times {10}^{3}J,
m=0.600kgat−12∘Cm=0.600kg\text{at}-12^{\circ}C,
S=2222.3J{kg}^{–1}^{\circ}{C}^{–1} and
L=336×103Jkg−1L=336\times {10}^{3}J{kg}^{-1}
Heat required to convert ice from –12∘C–12^{\circ}C to ice at 0∘C0^{\circ}C will be,
Q1=0.600×2222.3×12=16000.56J{Q}_{1}=0.600\times 2222.3\times 12=16000.56J
Remaining heat Q=184000−16000.56Q=184000-16000.56
=167999.44J=167999.44J
Heat required to melt ice completely at 0∘C0^{\circ}C,
Q2=0.600×336000=201600J{Q}_{2}=0.600\times 336000=201600J needed which is more than QQ, so 100100% ice is not melted.
For amount of melted ice, we can write
167999.44=m×336000⇒m=0.4999kg167999.44=m\times 336000 \Rightarrow m=0.4999kg
∴∴ mass of water =0.4999kg=0.4999kg
Mass of ice remaining =0.1001=0.1001
∴∴ Ratio =0.10010.4999≈1:5=\dfrac{0.1001}{0.4999}\approx 1:5

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