Gujarati
Hindi
10-2.Transmission of Heat
normal

Hot water cools from $60\,^oC$ to $50\,^oC$ in the first $10\, min$ and to $42\,^oC$ in the next $10\, min$. The temperature of the surrounding is  ........ $^oC$

A

$50$

B

$10$

C

$15$

D

$20$

Solution

$\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{t}}=\mathrm{k}\left(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}-\mathrm{T}_{0}\right)$

$\frac{60-50}{10}=\mathrm{k}\left(\frac{60+50}{2}-\mathrm{T}_{0}\right)$

$\frac{50-42}{10}=\mathrm{k}\left(\frac{50+42}{2}-\mathrm{T}_{0}\right)$

By solving $\mathrm{T}_{0}=10^{\circ} \mathrm{C}$

Standard 11
Physics

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