A sample of copper with a mass of 1.80 kg, initially at a temperature of 150.0°C, is in a well-insulated container. Water at a temperature of 27.0°C is added to the container, and the entire interior of the container is allowed to come to thermal equilibrium, where it reaches a final temperature of 70.0°C. What mass of water (in kg) was added? Assume any water turned to steam subsequently recondenses.

the mass of water is 0.3 Kg

Explanation:

since the container is well-insulated, the heat released by the copper is absorbed by the water, therefore:

Q water + Q copper = Q surroundings = 0 (insulated)

Q water = - Q copper

since Q = m * c * (T eq - Ti), where m = mass, c = specific heat, T eq = equilibrium temperature and Ti = initial temperature

and denoting w as water and co as copper:

m w * c w * (T eq - Tiw) = - m co * c co * (T eq - Ti co) = m co * c co * (T co - Ti eq)

m w = m co * c co * (T co - Ti eq) / [ c w * (T eq - Tiw) ]

We take the specific heat of water as c = 1 cal/g °C = 4.186 J/g °C. Also the specific heat of copper can be found in tables → at 25°C c co = 0.385 J/g°C

if we assume that both specific heats do not change during the process (or the change is insignificant)

m w = m co * c co * (T eq - Ti co) / [ c w * (T eq - Tiw) ]

m w = 1.80 kg * 0.385 J/g°C (150°C - 70°C) / (4.186 J/g°C (70°C - 27°C))

m w = 0.3 kg