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किसी बर्तन में भरे तप्त भोजन का ताप $2$ मिनट में $94^{\circ} \,C$ से $86^{\circ} \,C$ हो जाता है जबकि कक्ष-ताप $20^{\circ}\, C$ है। $71^{\circ} \,C$ से $69^{\circ}\, C$ तक ताप के गिरने में कितना समय लगेगा ?
$28$
$70$
$68$
$42$
Solution
Answer The average temperature of $94^{\circ} C$ and $86^{\circ} C$ is $90^{\circ} C ,$ which is $70^{\circ} C$ above the room temperature. Under these conditions the pan cools $8^{\circ} C$ in 2 minutes. , we have
$\frac{\text { Change in temperature }}{\text { Time }}=K \Delta T$
$\frac{8^{\circ} C }{2 mtn }=K\left(70^{\circ} C \right)$
The average of $69^{\circ} C$ and $71^{\circ} C$ is $70^{\circ} C ,$ which is $50^{\circ} C$ above room temperature. $K$ is the same for this situation as for the original.
$\frac{2^{\circ} C }{\text { Time }}=K\left(50^{\circ} C \right)$
When we divide above two equations, we have
$\frac{8^{\circ} C / 2 m 1 n }{2^{\circ} C / time }=\frac{K\left(70^{\circ} C \right)}{K\left(50^{\circ} C \right)}$
Time $=0.7 min$
$=42\, s$
