Thermodynamics help please?

Question

A 20 kg container of water having a temperature of 100°C is added to a 200 kg tub of water having a temperature of 10 °C. What is the final equilibrium temperature (in °C) of the mixture?

Answer 1

Q = m * c * ΔT

m = mass

c = specific heat capacity of water is 4.18 J/g ºC

ΔT = change in temperature

Q1 = -Q2 = thermal equilibrium

Q1 = 20 kg * 4.18 J/g ºC * (Tf - 100°C)

Q2 = 200 kg * 4.18 J/g ºC * (Tf - 10°C)

20 * 4.18 * (Tf - 100) = - (200 * 4.18 * (Tf - 10))

2 * ( Tf - 100 )  =  - (20 * ( Tf - 10 ))

2Tf - 200 = -20Tf + 200

2Tf + 20Tf = 200 + 200

Tf = 400 / 22 = 18 °C

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