`A` calorimeter contains `4.00` kg of water at `20.0^(@)C` . What amount of ice at `-10^(@)C` , must be added to cause the resulting mixture to rach thermal equilibruim at `5.0^(@)C` . Asuume that heat transfer occurs only between the water and ice.

Options

A. `0.55 kg`
B. `0.66 kg`
C. `2.51 kg`
D. `0.25 kg`

Answer

B

Explanation

Let m be the mas of ice needed.
`(("Heat of rise"),("m kg of ice"),("from" - 10^(0) "C to" 0^(0)))+(("Heat to change"),("m kg of ice"),("to water"))`
`+(("Heat of raise"),("m kg of water"),("from" - 10^(0) "C to" 5.0^(0)C))=(("Heat lost by4 kg"),("of water colling"),("from"20^(0) C "to" 5^(0)C))`
`m_(ice)C_(ice) Delta T+m_(ice)L_(f) +m_(ice)C_(w)Delta T`
`m(2100)[0^(0) C-(-10^(0)C)]+m(333)`
`+m(4186)(5.0^(0)-0.0^(0)C)`
`m(2100)(10)+333 m+m(210)=2510 kJ`.