Question
How much faster does a cup of coffee cool off from 100°C than from 30°C? Assume the coffee to act as a blackbody and temperature of surrounding is 20°C.
✓
Answer
In the first case, $\text{T}_1 = 100 ^o\text{C} =100 + 273 = 373 \text{K} $ In the second case. $\text{T}_2 = 30^o\text{C}= 30 +273= 303 \text{K }$ In both case, $\text{T}_1 = 20 ^o\text{C}= 20 + 273 = 293 \text{K}$ Using Stefan Boltzmann law,$\frac{\text{E}_1}{\text{E}_2}= \frac{\text{T}^4_1-\text{T}^4_1}{\text{T}^4_1-\text{T}^{4}_1}$
$= \frac{(373)^4-(293)^4}{(303)^4(293)^4}= \frac{1198}{105.88}$
$ = \frac{\text{E}_1}{\text{E}_2}= 11.3$
Cup of coffee cools off from than from about time faster.
$= \frac{(373)^4-(293)^4}{(303)^4(293)^4}= \frac{1198}{105.88}$
$ = \frac{\text{E}_1}{\text{E}_2}= 11.3$
Cup of coffee cools off from than from about time faster.
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