In Newton's experiment of cooling, water equivalent of two similar calorimeters is 10 gm each. T

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10-2.Transmission of Heat

easy

In Newton's experiment of cooling, the water equivalent of two similar calorimeters is $10 $ gm each. They are filled with $350 gm$ of water and $300 gm$ of a liquid (equal volumes) separately. The time taken by water and liquid to cool from ${70^o}C$ to ${60^o}C$ is $3$ min and $95$ sec respectively. The specific heat of the liquid will be ...... $Cal/gm\,^oC$

$0.3$

$0.5$

$0.6$

$0.8$

Solution

(c) ${S_l} = \frac{1}{{{m_l}}}\left[ {\frac{{{t_l}}}{{{t_W}}}({m_W}{C_W} + W) – W} \right]$
$ = \frac{1}{{300}}\left[ {\frac{{95}}{{3 \times 60}}(350 \times 1 + 10) – 10} \right]$ $=0.6 Cal/gm\times°C$

Standard 11

Physics

For a small temperature difference between the body and the surroundings the relation between the rate of loss heat $R$ and the temperature of the body is depicted by

medium

A cane is taken out from a refrigerator at $0°C$ . The atmospheric temperature is $25°C$ . If $t_1$ is the time taken to heat from $0°C$ to $5°C$ and $t_2$ is the time taken from $10°C$ to $15°C$, then

easy

A body takes $5$ minute to cool from $80°C$ to $50°C$ . …… $\min.$ it will take to cool from $60°C$ to $30°C$ , if room temperature is $20°C$ .

medium

Hot water cools from ${60^o}C$ to ${50^o}C$ in the first $10$ minutes and to ${42^o}C$ in the next $10$ minutes. The temperature of the surrounding is ……… $^oC$

hard

The temperature of a body falls from ${50^o}C$to ${40^o}C$ in $10$ minutes. If the temperature of the surroundings is ${20^o}C$ Then temperature of the body after another $10$ minutes will be …….. $^oC$