First slide

Heat transfer-Radiation

Question

Hot water cools from 60^oC to 50^oC in the first 10 minutes and to 42^oC in the next 10 minutes. The temperature of the surrounding is

Moderate

A

$5^{\circ} C$
B

$10^{\circ} C$
C

$15^{\circ} C$
D

$20^{\circ} C$

Solution

According to Newton's law of cooling,
$\frac{θ_{1} - θ_{2}}{t} = K [\frac{θ_{1} + θ_{2}}{2} - θ_{0}]$
In the first case, $\frac{(60 - 50)}{10} = K [\frac{60 + 50}{2} - θ_{0}]$
$1 = K (55 - θ_{0})$ ………..(i)
In the second case $\frac{(50 - 42)}{10} = K [\frac{50 + 42}{2} - θ_{0}]$
$0 .8 = K (46 - θ_{0})$ ………(ii)
Dividing (i) by (ii), we get $\frac{1}{0 .8} = \frac{55 - θ}{46 - θ}$
$or 46 - θ_{0} = 44 - 0 .8 θ_{0} \Rightarrow θ_{0} = 10^{\circ} C$

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