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When 0.1 kg of ice at 0^@ C is mixed wit...

When 0.1 kg of ice at `0^@ C` is mixed with 0.32 kg of water at `35^@ C` in a container. The resulting temperature of the mixture is `7.8^@ C`. Calculate the heat of fusion of ice (`s_("water") = 4186 J kg^(-1) K^(-1)`).

Text Solution

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Given
`m_("ice") = 0.1 kg`
` m_("water") = 0.32 kg`
` T_("ice") = 0^@ C`
` T_("ice") = 0^@ C `
`T_("Water ") = 35 ^@ C`
` T_(F) = 7.8^@ C `
` S_("water") = 4186 kg K^(-1)`
heat lost by water
` = m_("water ")s_("water") (T_(F) - T_("water"))`
`=0.32 kg xx 4186 j xx ( 7.8 - 35 )^@ C`
`=- 36434.944 J` ( here negative sign indicates loss of heat energy )
Heat required to melt ice ` = m_("ice") L_f = 0.1 xxL_f`
heat required to raise temperature of water ( from ice ) to final temperature
` = m_("ice" ) s (T - T_("ice"))`
` = 0.1 kg xx 4186 J xx ( 7.8 - 0 )C^@`
`= 3265.08 J`
head lost = heat gained
` 36434 .944 = 0.1 L_f + 3265.08`
` L_f = ( 36434 - 3265.08 )/( 0.1 )= 3316.9864`
`= 3.31698 xx 10^5 J kg^(-1)`
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