A metal cup has mass 0.12 kg and initial temperature 2 0 ∘ C . Into this cup is poured 0.30 kg of water (specific heat 4200 J/kg - ∘ C ), which is initially at temperature 7 0 ∘ C . The final temperature of the cup-water system is 6 6 ∘ C . Assuming no heat is lost to the surrounding, what is the specific heat of the cup?

Heat gained by cup = heat lost by water: 0.12 C ( 66 − 20 ) = 0.30 × 4200 × ( 70 − 66 ) 5.52 C = 5040 C = 23 21000 J/kg ∘ C

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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoCalorimetryMedium2 minQB

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A metal cup has mass $0.12 kg$ and initial temperature $2 0^{\circ} C$ . Into this cup is poured $0.30 kg$ of water (specific heat $4200 J/kg$ - $^{\circ} C$ ), which is initially at temperature $7 0^{\circ} C$ . The final temperature of the cup-water system is $6 6^{\circ} C$ . Assuming no heat is lost to the surrounding, what is the specific heat of the cup?

Options:

Answer:

Solution:

Heat gained by cup = heat lost by water:
$0.12 C (66 - 20) = 0.30 \times 4200 \times (70 - 66)$
$5.52 C = 5040$
$C = \frac{21000}{23} J/kg^{\circ} C$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Calorimetry

⏱ 2mℹ️ Source: QB

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