If you have 0.261 m3 of water at 25.0°C and add 0.127 m3 of water at 95.0°C, what is the final temperature of the mixture? Use 1000 kg/m3 as the density of water at any temperature.

Assuming that the specific heat capacity for both, we can rewrite the equation:

q = m * Cp * (T2 - T1) to q = m * (T2 - T1)

Equating them to each other we have:

m1 * (T2 - T1) = m2 * (T2 - T3)

0.261 * (T2 - 25) = 0.127 * (T2 - 95)

T2 = - 41.34 degrees Celsius