A steel vessel weighing 300 g is filled with 1.5 liters of water at 17 ° C. A 200 g lump of wet snow was dropped into the water.

A steel vessel weighing 300 g is filled with 1.5 liters of water at 17 ° C. A 200 g lump of wet snow was dropped into the water. When the snow melted, the temperature was set at 7 ° C. How much water was in the snow? The specific heat capacity of steel is 460J / kgK.

Let us write down the heat balance equation in order to solve the problem:
m1 * (T2 – T) * c_v + m2 * (L + (T2 – T) * c_v) + m3 * (T2 – T1) * c_s + m4 * (T2 – T1) * c_v = 0;
transform:
m1 * (T2 – T) * c_v + (m – m1) * (L + (T2 – T) * c_v) + m3 * (T2 – T1) * c_s + ro * V * (T2 -T1) * c_v = 0;
Let’s transform:
m1 * (T2 – T) * c_v – m1 * (L + (T2 – T) * c_v) = – m3 * (T2 – T1) * c_s – ro * V * (T2 – T1) * c_v – m * ( L + (T2 – T) * c_v);
Let’s transform:
m1 * ((T2 – T) * c_v – (L + (T2 – T) * c_v)) = – m3 * (T2 – T1) * c_s – ro * V * (T2 – T1) * c_v – m * ( L + (T2 – T) * c_v);
Let’s transform:
m1 = (- m3 * (T2 – T1) * c_s – ro * V * (T2 – T1) * c_v – m * (L + (T2 – T0) * c_v)) / ((T2 – T) * c_v – (L + (T2 – T) * c_v));
Let’s substitute and solve:
m1 = (- 0.3 * (7 – 17) * 460 – 1000 * 0.0015 * (7 – 17) * 4200 – 0.2 * (335000 + (7 – 0) * 4200)) / ((7 – 0) * 4200 – (335000 + (7 – 0) * 4200));
m1 = 0.025373134 kg;
m1 = 25.37 g;
m1 = 25 g;
Answer: m1 = 25 g.