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(B) The temperature is increased from $10^{\circ}C$ to $100^{\circ}C$. The total change in temperature is $\Delta T = 100^{\circ}C - 10^{\circ}C = 90^

Vedclass · Accepted Answer

$512$

Vedclass · Answer

(B) The temperature is increased from $10^{\circ}C$ to $100^{\circ}C$. The total change in temperature is $\Delta T = 100^{\circ}C - 10^{\circ}C = 90^{\circ}C$.Since the rate of emission doubles for every $10^{\circ}C$ increase,we calculate the number of such intervals as $n = \frac{90^{\circ}C}{10^{\circ}C} = 9$.Given that the rate of emission doubles for each interval,the new rate of emission relative to the initial rate is given by $2^n$.Substituting $n = 9$,we get $2^9 = 512$.Therefore,the rate of emission increases by a factor of $512$.

The rate of emission of a body doubles when its temperature is increased by $10^{\circ}C$. If the temperature of the body is increased from $10^{\circ}C$ to $100^{\circ}C$,the rate of emission increases by a factor of:

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