According to Newton’s law of cooling, the rate of cooling of a body is proportional to ${(\Delta \theta )^n}$, where $\Delta \theta $ is the difference of the temperature of the body and the surroundings, and n is equal to
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Liquid is filled in a vessel which is kept in a room with temperature ${20^o}C$. When the temperature of the liquid is ${80^o}C$, then it loses heat at the rate of $60\;cal/\sec $. What will be the rate of loss of heat when the temperature of the liquid is ${40^o}C$ ....... $cal/\sec $
A liquid cools down from ${70^o}C$ to ${60^o}C$ in $5$ minutes. The time taken to cool it from ${60^o}C$ to ${50^o}C$ will be
Read the following statements:
$A.$ When small temperature difference between a liquid and its surrounding is doubled the rate of loss of heat of the liquid becomes twice.
$B.$ Two bodies $P$ and $Q$ having equal surface areas are maintained at temperature $10^{\circ}\,C$ and $20^{\circ}\,C$. The thermal radiation emitted in a given time by $P$ and $Q$ are in the ratio $1: 1.15$
$C.$ A carnot Engine working between $100\,K$ and $400\,K$ has an efficiency of $75 \%$
$D.$ When small temperature difference between a liquid and its surrounding is quadrupled, the rate of loss of heat of the liquid becomes twice.
Choose the correct answer from the options given below :
A cup of tea cools from $65.5^o C to 62.5 ^o C$ in one minute in a room of $22.5 ^o C$. How long will the same cup of tea take, in.............. minutes, to cool from $46.50^o C to 40.5 ^o C$ in the same room ? (choose nearest value)
Hot water kept in a beaker placed in a room cools from ${70^o}C$ to $60°C$ in $4$ minutes. The time taken by it to cool from ${69^o}C$ to ${59^o}c$ will be