A 25.0 g piece of aluminum (which has a molar heat capacity of 24.03 J/ °C.mol) is heated 82.4 °C and dropped into a calorimeter containing water (specific heat capacity of water is 4.18 J/g.oC) initially at 22.3 °C. The final temperature of the water is 24.9 °C.Calculate the mass of water in the calorimeter.


117.64 g


Upon inspection, one of the given heat capacities is in J/°C mol; we will convert it to J/g°C

24.03\frac{\text{J}}{mol\text{ Al }\degree\text{C}}\frac{1\text{ mol Al}}{26.9815\text{ g Al}}=0.889498\text{ }\frac{\text{J}}{\text{ g}\degree\text{C}}

To solve this problem, we will assume that the heat released by the aluminum piece is the same as the heat absorbed by the water.

We have the specific heat formula

Q=mc\Delta T


Q=heat energy


c=specific heat capacity

Δ T=change in temperature

Then we will compare both equations and consider the variables, with subindex 1 being for water and 2 for aluminum

m_1c_1\Delta T_1=m_2c_2\Delta T_2

Solving for m1, we have

m_1=\frac{m_2c_2\Delta T_2}{c_1\Delta T_1}=\frac{-25\text{ g }0.8894\text{ J/ g}\degree\text{C}(24.9-82.4)\degree\text{C}}{4.18\text{ J/ g}\degree\text{C}(24.9-22.3)\text{ }\degree\text{C}}=117.64\text{ g}m_1=m_2c_2\Delta T_2=25\text{ g }0.8894\text{ J/ g}\degree\text{C}(24.9-82.4)\degree\text{C}=-1278.51\text{ g}

m_2=\frac{Q_2}{c_1\Delta T_1}=\frac{1278.51}{4.18\text{ J/ g}\degree\text{C }(24.9-22.3)}=117.64\text{ g}

Rate answer
Wrong answer?

If your question is not fully disclosed, then try using the search on the site and find other answers on the subject Chemistry.

Find another answers

Load image