Question

.Solve numerical. -2marks

m= 3000 gram, Q = 600 cal, change in temperature is equal to 10 degree centigrade ,find

specific heat in calorie per gram degree centigrade.​

Answer 1

Given :

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• $\sf Mass = m = 3000 gram$
• $\sf Q = 600 cal$
• $\sf Change \: in \: Temperature = \triangle T = 10^{\circ} C$

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To find :

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• $\sf Specific \: heat \:capacity = C = ??$

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Solution :

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$\bigstar\:{\underline{\sf We \:know \:that \::}}$

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$\star\;{\boxed{\sf{\pink{ Q = mc\triangle t}}}}$

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$:\implies\sf 600 = 3000\times c \times 10$

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$:\implies\sf c = \dfrac{600}{3000\times 10}$

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$: \implies \sf c = \dfrac{200}{1000\times 10}$

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$: \implies \sf c = \dfrac{20}{1000}$

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$: \implies \sf c = 0.02 cal \: gram^{-1} \: ^{\circ} C^{-1}$

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$:\implies{\underline{\boxed{\frak{\purple{ Specific \: heat \: Capacity = 0.02 cal\: gram^{-1} \: ^{\circ} C^{-1} }}}}}\;\bigstar$

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$\therefore\:{\underline{\sf{Specific \;heat \; capacity \; of \:the \; body \; is \bf{ 0.02 cal\: g^{-1} C^{-1} }.}}}$

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Answer 2

Given

• m = 3000 g
• Q = 600 cal
• Change in temperature = 10°C

To find

• Specific heat in cal/g°C

Solution

The quantitative relationship between heat transfer and temperature.

• Q = mcΔt

Where :-

• Q = 600 cal
• Δt = 10°C
• m = 3000

Substituting we get :-

• 600 = 3000 × c × 10
• c = 600 / 3000 × 10
• c = 200 / 1000 × 10
• c = 0.02

∴ The specific heat (c) is 0.02 cal/g°C