When 300 J of heat is added to 25 gm of sample of a material its temperature rises from $25^{\circ} C t o 45^{\circ} C .$ the thermal capacity of the sample and specific heat of the material are respectively given by

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Solution

The correct option is A
$T h e r m a l c a p a c i t y = m c = \frac{Q}{Δ T} = \frac{300}{45 - 25} = \frac{300}{20} = 15 J /^{\circ} C S p e c i f i c h e a t = \frac{T h e r m a l c a p a c i t y}{M a s s} = \frac{15}{25 \times 10^{- 3}} = 600 J / k g^{\circ} C$

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Q. When 300 J of heat is added to 25 gm of sample of a material its temperature rises from

25^{\circ} C t o 45^{\circ} C .

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When 300 J of heat is added to 25 gm of sample of a material its temperature rises from 25∘C to 45∘C. the thermal capacity of the sample and specific heat of the material are respectively given by

The correct option is A Thermal capacity=mc=QΔT=30045−25=30020=15J/∘CSpecific heat=Thermal capacityMass=1525×10−3=600J/kg∘C

When 300 J of heat is added to 25 gm of sample of a material its temperature rises from $25^{\circ} C t o 45^{\circ} C .$ the thermal capacity of the sample and specific heat of the material are respectively given by

The correct option is A
$T h e r m a l c a p a c i t y = m c = \frac{Q}{Δ T} = \frac{300}{45 - 25} = \frac{300}{20} = 15 J /^{\circ} C S p e c i f i c h e a t = \frac{T h e r m a l c a p a c i t y}{M a s s} = \frac{15}{25 \times 10^{- 3}} = 600 J / k g^{\circ} C$