1 kg metal at 10 0 ∘ C is cooled by 1 kg water at 0 ∘ C . If specific heat capacity of metal is 2 1 specific heat capacity of water, the final temperature is

Heat lost by metal = heat gained by water: 1 × 2 S × ( 100 − T ) = 1 × S × ( T − 0 ) 100 − T = 2 T T = 3 100 ∘ C This is less than 5 0 ∘ C .

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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoCalorimetryMedium2 minQB

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$1 kg$ metal at $10 0^{\circ} C$ is cooled by $1 kg$ water at $0^{\circ} C$ . If specific heat capacity of metal is $\frac{1}{2}$ specific heat capacity of water, the final temperature is

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Answer:

Solution:

Heat lost by metal = heat gained by water:
$1 \times \frac{S}{2} \times (100 - T) = 1 \times S \times (T - 0)$
$100 - T = 2 T$
$T = \frac{100}{3}^{\circ} C$
This is less than $5 0^{\circ} C$ .

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Calorimetry

⏱ 2mℹ️ Source: QB

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