A hot piece of iron of mass 20 g is dropped into a calorimeter of water equivalent 40 g containing 260 g of water at 200C. If the equilibrium temperature is 600C, the initial temperature of iron piece is(specific heat of iron = 1.2 cal g–1 0C–1)

Heat loss by hot body = Heat gained by clod body 20×1.2×t−60=260+4040⇒ t=5600C

A hot piece of iron of mass 20 g is dropped into a calorimeter of water equivalent 40 g containing 260 g of water at 200C. If the equilibrium temperature is 600C, the initial temperature of iron piece is(specific heat of iron = 1.2 cal g–1 0C–1)

Heat loss by hot body = Heat gained by clod body 20×1.2×t−60=260+4040⇒ t=5600C

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A hot piece of iron of mass 20 g is dropped into a calorimeter of water equivalent 40 g containing 260 g of water at 20⁰C. If the equilibrium temperature is 60⁰C, the initial temperature of iron piece is
(specific heat of iron = 1.2 cal g^–1 ⁰C^–1)

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560⁰C

420⁰C

500⁰C

440⁰C

answer is A.

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Detailed Solution

Heat loss by hot body = Heat gained by clod body

$20 \times 1.2 \times (t - 60) = (260 + 40) (40)$

$\Rightarrow t = 560^{0} C$

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A hot piece of iron of mass 20 g is dropped into a calorimeter of water equivalent 40 g containing 260 g of water at 200C. If the equilibrium temperature is 600C, the initial temperature of iron piece is(specific heat of iron = 1.2 cal g–1 0C–1)

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A hot piece of iron of mass 20 g is dropped into a calorimeter of water equivalent 40 g containing 260 g of water at 20⁰C. If the equilibrium temperature is 60⁰C, the initial temperature of iron piece is
(specific heat of iron = 1.2 cal g^–1 ⁰C^–1)