COMEDK · Physics · 12. Thermal Properties of Matter
A pan filled with hot food cools from \(94^{\circ} \mathrm{C}\) to \(86^{\circ} \mathrm{C}\) in \(2 \mathrm{~min}\), when the room temperature is \(20^{\circ} \mathrm{C}\). The time taken for the food to cool from \(86^{\circ} \mathrm{C}\) to \(74^{\circ} \mathrm{C}\) will be
- A \(500 \mathrm{~s}\)
- B \(420 \mathrm{~s}\)
- C \(200 \mathrm{~s}\)
- D \(210 \mathrm{~s}\)
Answer & Solution
Correct Answer
(D) \(210 \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
According to Newton's law of cooling, \[ \frac{T_{1}-T_{2}}{t}=K\left(\frac{T_{1}+T_{2}}{2}-T_{s}\right) ...(i) \] where, \(T_{s}\) is the temperature of surrounding. For the first case, \(T_{1}=94^{\circ} \mathrm{C}, T_{2}=86^{\circ} \mathrm{C}\)…
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