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Standard X

Physics

Principle of Calorimetry

20g of ice an...

Question

$20 g$ of ice and $20 g$ of hot water are mixed, when the ice is melted the temperature of the mixture was found to be $0^{o} C$ . The temperature of hot water taken should be
( $L_{i c e} =$ 80 cal/g)

40^{0} C

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72^{0} C

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80^{0} C

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96^{0} C

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Solution

The correct option is C $80^{0} C$
Assuming that the ice was at 0 $^{o}$ C initially,
Heat lost by the hot water $=$ Heat gained by the ice
$m_{1} . c_{1} . Δ T_{1} = m_{2} . L_{i}$ [since the final temperature of mixture is 0 $^{o}$ C]
$20 (1) (T - 0) = 20 (80)$
$T = 80^{o} C$

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20g of ice and 20g of hot water are mixed, when the ice is melted the temperature of the mixture was found to be 0oC. The temperature of hot water taken should be (Lice=80 cal/g)

The correct option is C 800CAssuming that the ice was at 0oC initially,Heat lost by the hot water=Heat gained by the icem1.c1.ΔT1=m2.Li[since the final temperature of mixture is 0oC]20(1)(T−0)=20(80)T=80oC

$20 g$ of ice and $20 g$ of hot water are mixed, when the ice is melted the temperature of the mixture was found to be $0^{o} C$ . The temperature of hot water taken should be
( $L_{i c e} =$ 80 cal/g)

The correct option is C $80^{0} C$
Assuming that the ice was at 0 $^{o}$ C initially,
Heat lost by the hot water $=$ Heat gained by the ice
$m_{1} . c_{1} . Δ T_{1} = m_{2} . L_{i}$ [since the final temperature of mixture is 0 $^{o}$ C]
$20 (1) (T - 0) = 20 (80)$
$T = 80^{o} C$