Calorimetry/Thermodynamics problem. Please help if you can!

Question

If you have 0.266 m³ of water at 25°C in an insulated container and add 0.112 m³ of water at 95°C, what is the final temperature T_f of the mixture? Use 1000 kg/m³ as the density of water at any temperature. 

Answer 1

heat lost by hot water = heat gained by cool water

heat lost or gained = mass x specific heat x change in temperature = mC∆T

First, let's convert m3 to liters and then to mass (g or kg)

0.266 m³ x 1000 L/m³ x 1 kg/L = 266 kg cool water

0.112 m³ x 1000 L/m³ x 1 kg/L = 112 kg hot water

(I just noticed they gave the conversion factor of 1000 kg/m³), so you could have done this...

0.266 m³ x 1000 kg/m3 = 266 kg, etc.

Nonetheless, to continue...

heat from hot water = mC∆T = (112 kg)(4.184 kJ/kgº)(95º-Tf)

heat to cool water = mC∆T = (266kg)4.184 kJ/kgº)(Tf - 25º)

Setting these equal to each other we have:

(112 kg)(4.184 kJ/kgº)(95º-Tf) = (266kg)(4.184 kJ/kgº)(Tf - 25º)

44518 - 469Tf = 1113Tf - 27824

1582Tf = 71802

Tf = 45.4ºC