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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoCalorimetryMedium2 minQB

PhysicsMediumsingle choice

Passage / Comprehension

Given a liquid initially at -20°C, warmed in a 2520 W microwave oven. Its mass is 1 kg and specific heat s = 2(T - 253) cal/(g K), where T is in kelvin. Take T = θ + 273, where θ is in °C.

Determine the temperature vs. time function

Options:

Answer:

Solution:

$P = \frac{dQ}{dt}$ $Pdt = dQ \int_{0}^{t} P dt = \int_{- 20}^{θ} msd θ \int_{0}^{t} 2520 dt = \int_{- 20}^{θ} 1 \times 2 (θ^{\circ} + 273 - 253) \times 4200 d θ$ $2520 t = \int_{- 20}^{θ} 2 (θ^{\circ} + 20) \times 4200 d θ \frac{2520}{8400} t = [θ^{2} + 20 θ]_{- 20}^{θ} \frac{3 t}{5} = θ^{2} + 40 θ + 400 3 t = 5 θ^{2} + 200 θ + 2000$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Calorimetry

⏱ 2mℹ️ Source: QB

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