50 g of ice at 0°C is added to 300 g of a liquid at 30°C. What will be the finaltemperature of the mixture when all the ice has melted? The specific heat capacity ofthe liquid is 265Jg-1°C-1 while that of water is 4 2 Jg-1°C-1.Specific latent heat of fusion of ice =336Jg-1 .[2]

To solve this problem, we need to consider the energy balance in the system. The energy gained by the ice as it melts and warms up to the final temperature must be equal to the energy lost by the liquid as it cools down to the final temperature.

1. First, calculate the energy required to melt the ice. This is given by the mass of the ice times the specific latent heat of fusion of ice:

   Energy to melt ice = mass_ice * latent_heat_ice = 50g * 336 J/g = 16800 J

2. The ice will then warm up to the final temperature. We don't know this temperature yet, but we can call it T. The energy required for this is given by the mass of the ice times the specific heat capacity of water times the change in temperature:

   Energy to warm ice = mass_ice * specific_heat_water * (T - 0°C) = 50g * 4.2 J/g°C * T = 210T J

3. The liquid will cool down to the final temperature T. The energy lost by the liquid is given by the mass of the liquid times the specific heat capacity of the liquid times the change in temperature:

   Energy lost by liquid = mass_liquid * specific_heat_liquid * (30°C - T) = 300g * 2.65 J/g°C * (30°C - T) = 79500 - 795T J

4. The total energy in the system remains constant, so the energy gained by the ice is equal to the energy lost by the liquid:

   16800 J + 210T J = 79500 J - 795T J

5. Solving this equation for T gives:

   1005T = 62700 T = 62.4°C

So, the final temperature of the mixture when all the ice has melted is 62.4°C.