Question

A piece of ice of mass 60 g is dropped into 140 g of water at 50 °c when all the ice has melted.
(Assume no heat is lost to the surrounding)
Specific heat capacity of water=4.2Jgk
Specific latent heat fusion of ice=336Jg
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Answer 1

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Explanation:

Answer 2

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Let the final temperature of water be T .

Thermal energy gained by ice = $m_{ice}\times ( L + C_{p} T )$

$= 60 \times ( 336 + 4.2 × T )$

$Where\:m_{ice}\:is\:mass\:of\:ice , L\:is\:latent\\heat\:of\:fusion\:of\:ice\\and\:C_{p}\:is\:specific\:heat\:of\:water$

$Thermal\:energy\:lost\:by\:water \\= m_{w}\times C_{p}\times ( 50 - T ) \\ = 140 \times 4.2 \times (50-T)$

$where\:m_{w}\:is\:mass\:of\:water$

By law of conservation of energy ,

we get ,

140 × 4.2 × ( 50 - T ) → 60 × ( 336 + 4.2 × T )

By solving this we get ,

$\huge{\mathfrak{T = 11° C }}$

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