Question

a calorimeter contains 150 g of water at 22.0°c. a hot metal object is placed in the water, and the final temperature of the water and metal becomes 28.5°c. if the water absorbed 4087.5 j of heat, what was the mass of the metal object if its specific heat capacity is 0.50 j/g°c and its initial temperature was 100.0°c?
50.0 g
110.0 g
150.0 g
100.0 g

Explanation:

Step1: Use heat - transfer formula

The heat absorbed by water is equal to the heat released by the metal. The formula for heat transfer is \(Q = mc\Delta T\), where \(Q\) is the heat, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the temperature change. For the metal, \(Q_{metal}=-Q_{water}\) (negative because the metal loses heat). We know \(Q_{water} = 4087.5\space J\), \(c_{metal}=0.50\space J/g^{\circ}C\), \(T_{i - metal}=100.0^{\circ}C\), \(T_{f}=28.5^{\circ}C\). Then \(\Delta T_{metal}=T_{f}-T_{i - metal}=28.5 - 100.0=-71.5^{\circ}C\)

Step2: Solve for the mass of the metal

From \(Q = mc\Delta T\), we can solve for \(m\). Rearranging the formula for \(m\) gives \(m=\frac{Q}{c\Delta T}\). Substituting \(Q = - 4087.5\space J\) (since \(Q_{metal}=-Q_{water}\)), \(c = 0.50\space J/g^{\circ}C\) and \(\Delta T=-71.5^{\circ}C\) into the formula:

$$m=\frac{-4087.5}{0.50\times(- 71.5)}=\frac{-4087.5}{-35.75}=114.33\approx110.0\space g$$

Answer:

110.0 g