M води=300г
t1=10°C
M льоду=400г
t2=-20°С
t3=0°C
h(лямбда)-330000Дж/кг
C води=4190 Дж/кг•°С
С льоду=2100 Дж/кг•°С
r-?

Предыдущий
вопрос Следующий
вопрос

To calculate the specific heat capacity of the water-ice system, we can use the principle of conservation of energy. The heat lost by the water at 10°C (M_water C_water (t3 - t1)) equals the heat gained by the ice at -20°C (M_ice C_ice (t3 - t2)).

Let's plug in the values:

M_water = 300g = 0.3 kg
t1 = 10°C
M_ice = 400g = 0.4 kg
t2 = -20°C
t3 = 0°C
h = 330000 J/kg
C_water = 4190 J/kg°C
C_ice = 2100 J/kg°C

The heat lost = heat gained
0.3 4190 (0 - 10) = 0.4 2100 (0 - (-20)) + 0.4 * h

Calculating:

-12570 J = 16800 J + 0.4 * 330000
-12570 J = 16800 J + 132000 J
-12570 J = 148800 J
148800 J - 12570 J = 136230 J

Now, we know that the heat gained by the ice to raise the temperature from -20°C to 0°C and melt is equal to 136230 J. So:

136230 J = 0.4 2100 20 + 0.4 330000 + 0.4 334000
136230 J = 16800 J + 132000 J + 136000 J
136230 J = 285800 J

Therefore, the specific heat capacity of the system is 285800 J / 20°C = 14290 J/kg°C.