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.

Answer

117.64 g

Procedure

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

$Q=mc\Delta T$

Where

Q=heat energy

m=mass

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

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

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

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