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19 g of water at 30^@C and 5 g of ice at...

19 g of water at `30^@C` and 5 g of ice at `-20^@C` are mixed together in a calorimeter. What is the final temperature of the mixture? Given specific heat of ice `=0.5calg^(-1) (.^(@)C)^(-1)` and latent heat of fusion of ice `=80calg^(-1)`

A

`0^@ C `

B

`-5^(@)`C

C

`5^@ C `

D

`10^@ C`

Text Solution

Verified by Experts

The correct Answer is:
C
here ,specific heat of ice `S_(ice )- 0.5 cal g^(-1) ""^(@) C^(-1)`
specific heat of water `s_("water ") = 1 cal g^(-1 @) C^(-1)`
Latent heat of fusion of ice `L_(f _( ice))` = 80 cal `g^(-1)`

here ice will absorb heat while hot water will release it Let `T ` be the final temperature of the mixture .
Assuming water equivalent of calorimeter to be neglected heat given by water `Q_1= m_( "Water")s_("water") Delta T `
`=19 xx 1 xx (30 - T )= 570 - 19 T `
heat absorbed by ice
`Q_2 = m_(ice) xx s_(ice) xx [0-(-20)]+m_(ice) +l_(f ice ) +m_(ice)xx s_("water") xx (T-0)`
`= 5 xx 0.5 xx 20 +5 xx 80+5 xx 1xxT `
`Q_2 = 450 +5T `
According to the principle of calorimetry `Q_1 =Q_2`
using eq (i ) and (ii )
` i.e., 570 - 19 T = 450 + 5T or T = ( 120)/(24 ) = 5^@ C `
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