Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoRadiationMedium2 minQB

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Passage / Comprehension

By varying the voltage applied to the kettle, you can change its power consumption P. Depending on P, the kettle with water can be heated to different maximum temperatures. This dependence is shown in Table 1.

Table 1

Power P (W)	0	100	200	300
Temperature T (°C)	20	40	60	80

Table 2 shows the variation of temperature with time when the kettle with a 300 W heater is switched off. The heat capacity of the empty kettle C₀ = 100 J/K; specific heat of water = 4200 J/(kg K); density of water = 1000 kg/m³. Assume Newton’s law of cooling.

Table 2

Time t (s)	0	60	300	600	1200	2400
Temperature T (°C)	80	75	60	45	30	20

How many litres of water is present in the kettle?

Options:

Answer:

Solution:

For $1^{st} 60 sec$ , we can use approx method

$\frac{dT}{dt} = \frac{k}{ms + C _{0}} (Ta_{v} - T_{0}); \frac{5}{60} = \frac{5}{m s + 100} [77.5 - 20] \Rightarrow ms + 100 = 3450; m \times 4200 = 3350; m = \frac{3350}{4200} = 0.798 kg vol. = \frac{m}{p} = \frac{0.798}{1000} m^{3} = 0.798 litre$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Radiation

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