Mixed 3 liquids of the same mass and heat capacity, but different temperatures.

Mixed 3 liquids of the same mass and heat capacity, but different temperatures. the first has a temperature of 300K, the second 280K, the third 335K. determine the steady-state temperature.

Given:

m is the mass of liquids;

c – heat capacity of liquids;

T1 = 300 degrees Kelvin – the temperature of the first liquid;

T2 = 280 degrees Kelvin – the temperature of the second liquid;

T3 = 335 degrees Kelvin – the temperature of the third liquid.

It is required to determine the steady-state temperature of the mixture T (degree Kelvin).

From the condition of the problem it is seen that the highest temperature has liquid No. 3. This means that the first and second liquid will heat up, and the third will cool down. Then:

Q1 + Q2 = Q3;

c * m * (T – T1) + c * m * (T – T2) = c * m * (T3 – T)

T – T1 + T – T2 = T3 – T;

2 * T – T1 – T2 = T3 – T;

3 * T = T1 + T2 + T3;

T = (T1 + T2 + T3) / 3 = (300 + 280 + 335) / 3 = 915/3 = 305 Kelvin.

Answer: the temperature of the mixture will be equal to 305 Kelvin.