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Standard X

Physics

Principle of Calorimetry

Consider two ...

Question

Consider two hot bodies $B_{1}$ and $B_{2}$ which have temperatures $100^{o} C$ and $80^{o} C$ respectively at t = 0. The temperature of the surroundings is $40^{o} C$ . The ratio of the respective rates of cooling $R_{1}$ and $R_{2}$ , of these two bodies at t = 0 will be

R_{1} : R_{2} = 3 : 2

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R_{1} : R_{2} = 5 : 4

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R_{1} : R_{2} = 2 : 3

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R_{1} : R_{2} = 4 : 5

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Solution

The correct option is A $R_{1} : R_{2} = 3 : 2$
Initially at $t = 0$ , rate of cooling ( $R$ ) $\propto$ Fall in temperature of body $(θ - θ_{0})$
$\Rightarrow \frac{R_{1}}{R_{2}} = \frac{θ_{1} - θ_{0}}{θ_{2} - θ_{0}} = \frac{100 - 40}{80 - 40} = \frac{3}{2}$

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Consider two hot bodies B1 and B2 which have temperatures 100oC and 80oC respectively at t = 0. The temperature of the surroundings is 40oC. The ratio of the respective rates of cooling R1 and R2, of these two bodies at t = 0 will be

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Consider two hot bodies $B_{1}$ and $B_{2}$ which have temperatures $100^{o} C$ and $80^{o} C$ respectively at t = 0. The temperature of the surroundings is $40^{o} C$ . The ratio of the respective rates of cooling $R_{1}$ and $R_{2}$ , of these two bodies at t = 0 will be

The correct option is A $R_{1} : R_{2} = 3 : 2$
Initially at $t = 0$ , rate of cooling ( $R$ ) $\propto$ Fall in temperature of body $(θ - θ_{0})$
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