Question

Please
Help question 5.
0.5 kilogram of water at 50°C is kept in open its temperature falls to 40°C .How much heat is lost to the surrounding? Specific heat of the water is 4200 J/kg °C.

Answer 1

We have mixed 540 g of ice at 0°c with 540 g of water at 80°c. 

Given that 
latent heat of fusion = 80 cal/g 
sp. heat capacity of water =1 cal/g/°c 
sp. heat capacity of ice =0.5 cal/g/°c 

So first thing in this mixing process will occur, is heat transfer from water at 80°c to saturated ice at 0°c till thermodynamic equilibrium occurs.
Since ice is saturated, so it will absorb heat in the form of latent heat to convert into water.
Let final temp.( equilibrium) of mixture is = T°c
So energy balance equation,

Latent heat to ice at 0°c to convert into saturated water at 0°c + sensible heating of this saturated water from 0°c to final temp T°c = cooling of hot water from 80°c to final temp T°c

=> (540*80) + {540*1*(T-0)} = {540*1*(80-T)}

=> 540*T = -540*T

=> 1080*T = 0

=> T = 0

So final temp of mixture is 0°c and mixture is saturated water.

Answer 2

m=0•5kg
initial temperature = 50 ℃
final = 40°C
∆ T = T final - T initial
= 40 ℃ ➖ 50 ℃
∆T= -10°C
Q=m X ∆T x s
=0•5×(-10)×4200
=-21000J