Question

A 0.0500-kg ice cube at −30.0°C is placed in 0.450 kg of 35.0°C water in a very well-insulated container. What is the final temperature in degrees Celsius?

Answer 1

Given

   mass of the ice cubes is m1 = 0.0500 kg
   initial temperature is T1 = -30.0 0C

mass of water i s m2 = 0.450 kg
initial Temperature of the water is T2 = 35.0 0C

final temperature of the system is T = ?

here the heat energy lost by the ice cubes = heat energy gain by the water
and the ice will change the state at 0 0C so

the equation can be written as

   m1*C1*(0-(-30))+m1*L+m1*C2(T-0) = m2*C2*(T-T2)
here C1 is specific heat of ice C1 = 2010 J/kg.k

latent heat of fusion of ice is L = 333*10^3 J/kg

   specific heat of water is C2 = 4184 J/kg.k

so
   0.0500*2010(30)+0.0500*333*10^3+0.0500*4184(T-0) = 0.450*4184(35-T)

solving for T , T = 22.1 0C