Question

A piece of ice of mass 40g is added to 200g of water at 50 degree Celsius. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water =4200 J per kg per K and specific latent heat of fusion of ice=336*1000 J per kg.

Answer 1

$Toconvert{0}^{o}iceinto{0}^{o}water,heatrequiredbytheicesample$

$Q_1=mL$

$=40\times {10}^{-3}\times 336\times {10}^3$

$=13440J$

$Nowle{t}^{\prime }senquireifthismuchheatisavailablewithwaterornot?$

$so$

$Q_2=ms\Delta \theta$

$=200\times {10}^{-3}\times 4200(50-t)J$

$soifheatfromwaterconvertsiceintowateratsometemperaturethen,$

$Q_1+Q=Q_2$

$13440+m_{1}s\Delta \theta =200\times {10}^{-3}\times 4200(50-t)$

$13440+40\times {10}^{-3}\times 4200\left(t\right)=200\times {10}^{-3}\times 4200(50-t)$

13440 + 168t= 42000 – 840t

168t + 840t = 42000 – 13440

1008t = 28560

t = 28560/1008 = 28.33oC

Hence, the final temperature of water is 28.33oC