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Q. A can is taken out from a refrigerator at $0^{\circ} C$. The atmospheric temperature is $25^{\circ}C$. If $t_1$ is the time taken to heat from $0^{\circ} C$ to $5^{\circ} C$ and $t_2$ is the time taken from $10^{\circ} C$ to $15^{\circ}C$, then :

BITSATBITSAT 2006

Solution:

Sir Newton's law of heating states that rate of change of temperature of a object is proportional to the difference in temperatures of the surrounding and the object. In the given cases, when initial temperature of can is $0^{\circ} C$, the difference in temperatures of can and surrounding is $25-0=25^{\circ} C$.
When initial temperature of can is $10^{\circ} C$, the difference in temperatures of can and surrounding is $25-10=15^{\circ} C$ therefore the rate of heating will be higher, when temperature difference is $25^{\circ} C$ i.e. time $t_{1}$ will be smaller than
$t _{1}< t _{2}$