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Standard XII

Physics

Heat Capacity

An iron block...

Question

An iron block of mass $2 k g$ , falls from a height of $10 m$ . After colliding with the ground it loses $25 %$ energy to surroundings and rest is gained as heat. Then find the temperature rise of the block. (Take sp. heat of iron $470 J / k g^{\circ} C$ )

{0.53}^{\circ} C

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{0.053}^{\circ} C

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{0.159}^{\circ} C

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{0.212}^{\circ} C

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Solution

The correct option is D ${0.159}^{\circ} C$
$P E = 2 g \times 10 = 20 g$
Heat loss is $25 %$
So, remaining heat is $15 g$
$15 g = m c Δ T$
$\Rightarrow Δ T = {0.159}^{o} C$

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An iron block of mass 2kg, falls from a height of 10m. After colliding with the ground it loses 25% energy to surroundings and rest is gained as heat. Then find the temperature rise of the block. (Take sp. heat of iron 470J/kg∘C)

The correct option is D 0.159∘CPE=2g×10=20gHeat loss is 25%So, remaining heat is 15g15g=mcΔT⇒ΔT=0.159oC

An iron block of mass $2 k g$ , falls from a height of $10 m$ . After colliding with the ground it loses $25 %$ energy to surroundings and rest is gained as heat. Then find the temperature rise of the block. (Take sp. heat of iron $470 J / k g^{\circ} C$ )

The correct option is D ${0.159}^{\circ} C$
$P E = 2 g \times 10 = 20 g$
Heat loss is $25 %$
So, remaining heat is $15 g$
$15 g = m c Δ T$
$\Rightarrow Δ T = {0.159}^{o} C$