TS EAMCET · Physics · Thermal Properties of Matter
A hot body placed in air cools down to a lower temperature. The rate of decrease of temperature is proportional to the temperature difference from the surrounding. The body loses \(60 \%\) and \(80 \%\) of maximum heat it can loose in time \(t_1\) and \(t_2\) respectively. The ratio \(t_2 / t_1\) will be
- A \(\frac{\ln (10)}{\ln (2)}\)
- B \(\frac{\ln (8)}{\ln (6)}\)
- C \(\frac{\ln (1)}{\ln (3)}\)
- D \(\frac{\ln (5)}{\ln \left(\frac{5}{2}\right)}\)
Answer & Solution
Correct Answer
(D) \(\frac{\ln (5)}{\ln \left(\frac{5}{2}\right)}\)
Step-by-step Solution
Detailed explanation
According to Newton's law of cooling, \[ -\frac{d T}{d t} \alpha \Delta \mathrm{T} \text { or } \frac{d T}{d t}=-K \Delta \mathrm{T} \] Solved the above Eq., we get…
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