Ice having a temperature of 0 ° C was dropped into a bath containing 100 kg of water at a temperature of 80 ° C.

Ice having a temperature of 0 ° C was dropped into a bath containing 100 kg of water at a temperature of 80 ° C. After establishing thermal equilibrium, the water temperature dropped to 47 ° C. Ignoring heat losses, determine how many kilograms of ice were taken to cool the water?

Given:
m1 = 100kg,
t1 = 80 ° С,
t2 = 0 ° С,
t3 = 47 ° С,
c = 4200J / (kg * deg),
λ = 3.3 * 10 ^ 5 J / kg;
To find:
m2 -?
Let’s write the heat balance equation:
Q1 = Q2 + Q3, where
Q1 = m1 * c * (t1 – t3) – the amount of heat given off by the water in the bath;
Q2 = m2 * λ – the amount of heat spent on ice melting;
Q3 = m2 * c * (t3 – t2) amount of heat for heating water from melted ice;
m1 * c * (t1 – t3) = m2 * λ + m2 * c * (t3 – t2) = m2 * [c * (t3 – t2) + λ];
m2 = [m1 * c * (t1 – t3)] / [c * (t3 – t2) + λ];
m2 = (100 * 4200 * 33) / (3.3 * 10 ^ 5 + 4200 * 47) = (138.6 * 10 ^ 5) / (5.274 * 10 ^ 5);
m2 = 26.28kg.