Thermal Properties of Matter

JEE Main level, single correct, medium

53 of 126

Question 53 of 126

A gun fires a lead bullet of temperature 300 K into a wooden block. The bullet having melting temperature of 600 K penetrates into the block and melts down. If the total heat required for the process is 625 J , then the mass of the bullet is \qquad grams.
(Latent heat of fusion of lead =2.5×104JKg−1=2.5 \times 10^4 \mathrm{JKg}^{-1} and specific heat capacity of lead =125JKg−1=125 \mathrm{JKg}^{-1} K−1)\left.\mathrm{K}^{-1}\right)

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Answer: (A) 1010

625=msΔT+mL625=m[125×300+2.5×104]625=m[37500+25000]625=m[62500]m=1100 kgM=10 grams \begin{aligned} & 625=\mathrm{ms} \Delta \mathrm{T}+\mathrm{mL} \\ & 625=\mathrm{m}\left[125 \times 300+2.5 \times 10^4\right] \\ & 625=\mathrm{m}[37500+25000] \\ & 625=\mathrm{m}[62500] \\ & \mathrm{m}=\frac{1}{100} \mathrm{~kg} \\ & \mathrm{M}=10 \text { grams }\end{aligned}

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