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A calorimeter of heat capacity 83.72 J k...

A calorimeter of heat capacity 83.72 J `k^(-1)` contains 0.48 kg of water at `35^(@)`C. How much mass of ice at `0^(0)` C should be added to decrease m the temperature of the calorimeter to `20^(@)` C .

Text Solution

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Let .m. be the mass of ice added,
Heat gained by ice during melting = mL
= m` xx 0.335 xx 10^(6)` J
Heat gainerd by molten ice (water) during rise in temperature = m ` xx 4186 xx (20 - 0 ) = m xx 83720` J
`therefore ` Heat gained by the ice= Heat lost by the calorimeter and water
`m xx [ 0.335 xx 10^(6) + 83720 ] = 1255.8 + 30139.2 = 31395 J m `
m [ 33500 + 83720 ] = 31395
`m = (31395)/(418720) = `0.07498 Kg.
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