A freezer has a coefficient of performance 5. When 3.6 × 1 0 6 J of work is done on the freezer, what mass of water at 0 ∘ C is converted into ice at 0 ∘ C?

Heat removed = 5 × 3.6 × 1 0 6 = 1.8 × 1 0 7 J. m = 3.34 × 1 0 5 1.8 × 1 0 7 = 53.9 kg. C.

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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoThermodynamics Laws & Specific Heats of GasesMedium2 minQB

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A freezer has a coefficient of performance 5. When $3.6 \times 1 0^{6}$ J of work is done on the freezer, what mass of water at $0^{\circ}$ C is converted into ice at $0^{\circ}$ C?

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Answer:

Solution:

Heat removed = $5 \times 3.6 \times 1 0^{6} = 1.8 \times 1 0^{7}$ J. $m = \frac{1.8 \times 1 0 ^{7}}{3.34 \times 1 0 ^{5}} = 53.9$ kg. C.

Subject:PhysicsTopic:Heat & ThermoSubtopic:Thermodynamics Laws & Specific Heats of Gases

⏱ 2mℹ️ Source: QB

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