JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
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 __________ grams.
(Latent heat of fusion of lead \(=2.5 \times 10^4 \mathrm{JKg}^{-1}\) and specific heat capacity of lead \(=125 \mathrm{JKg}^{-1}\) \(\left.\mathrm{K}^{-1}\right)\)
- A \(10\)
- B \(20\)
- C \(5\)
- D \(15\)
Answer & Solution
Correct Answer
(A) \(10\)
Step-by-step Solution
Detailed explanation
\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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