After the tea is added to the thermos, the temperature of the liquid quickly falls from 80 ∘ C to 75 ∘ C as it reaches thermal equilibrium with the thermos flask. What is the heat capacity of the thermos?

Heat lost by tea equals heat gained by the thermos, initially at 20 ∘ C . 250 ( 4.18 ) ( 80 − 75 ) = C ( 75 − 20 ) C = 55 5225 C = 95 J/K

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Passage / Comprehension

A $500 g$ teapot and an insulated thermos are in a $20^{\circ} C$ room. The teapot is filled with $1000 g$ of boiling water. 12 tea bags are then placed into the teapot. The brewed tea is allowed to cool to $80^{\circ} C$ , then $250 g$ of the tea is poured from the teapot into the thermos. The teapot is then kept on an insulated warmer that transfers $500 cal$ /min to the tea. Assume that the specific heat of brewed tea is the same as that of pure water, and that the tea bags have a very small mass compared to that of the water, and a negligible effect on the temperature. The specific heat of teapot is $0.17 J/g$ K and that of water is $4.18 J/g$ K. The entire procedure is done under atmospheric pressure. There are $4.18 J$ in one calorie.

After the tea is added to the thermos, the temperature of the liquid quickly falls from $80^{\circ} C$ to $75^{\circ} C$ as it reaches thermal equilibrium with the thermos flask. What is the heat capacity of the thermos?

Options:

Answer:

Solution:

Heat lost by tea equals heat gained by the thermos, initially at $20^{\circ} C$ .
$250 (4.18) (80 - 75) = C (75 - 20)$
$C = \frac{5225}{55}$
$C = 95 J/K$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Calorimetry

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