200 g of water was poured into the calorimeter at t⁰ 30 ⁰C and 300 g of water at t⁰ 20 ⁰C after the establishment

200 g of water was poured into the calorimeter at t⁰ 30 ⁰C and 300 g of water at t⁰ 20 ⁰C after the establishment of thermal equilibrium, t⁰ of water in the calorimeter will be equal to …

Initial data: m1 (mass of water at a temperature of t1 = 30 ºC) = 200 g or 0.2 kg; m2 (mass of water at a temperature of t1 = 20 ºC) = 300 g or 0.3 kg.

Let’s draw up an equality to determine the steady-state temperature (t3) and calculate it:

C * m1 * (t1 – t3) = C * m2 * (t3 – t2).

m1 * (t1 – t3) = m2 * (t3 – t2).

0.2 * (30 – t3) = 0.3 * (t3 – 20).

6 – 0.2t3 = 0.3t3 – 6.

12 = 0.5t3.

t3 = 12 / 0.5 = 24 ºC.

Answer: After establishing thermal equilibrium, the temperature in the calorimeter will be equal to 24 ºC.