Answer:
0° C
Explanation:
Given that
Mass of ice, m = 50g
Mass of water, m(w) = 50g
Temperature of ice, T(i) = 0° C
Temperature of water, T(w) = 80° C
Also, it is known that
Specific heat of water, c = 1 cal/g/°C
Latent heat of ice, L(w) = 89 cal/g
Let us assume T to be the final temperature of mixture.
This makes the energy balance equation:
Heat gained by ice to change itself into water + heat gained by melted ice(water) to raise its temperature at T° C = heat lost by water to reach at T° C
m(i).L(i) + m(i).c(w)[T - 0] = m(w).c(w)[80 - T], on substituting, we have
50 * 80 + 50 * 1(T - 0) = 50 * 1(80 - T)
4000 + 50T = 4000 - 50T
0 = 100 T
T = 0° C
Thus, the final temperature is 0° C
