Question

exam 3 fall 2025
your initials
(iii) (15 points) if 1.0 g of sulfuric acid were actually formed in the reaction, then what was the percent yield?

1. (15 points) 54.80 g of an unknown metal is put into a calorimeter (see sketch in figure 4) that contains 100.0g of water. the metal sample starts off at 100.0°c and the temperature of the water starts off at 21.0°c. when the temperature of the water stops changing its 29.3°c. the specific heat of the water is 4.184 j/(g°c). the pressure remains constant at 1 atm.

calculate the specific heat of the unknown metal sample.
figure 4: a coffee cup calorimeter.

Explanation:

Step1: Calculate heat absorbed by water

Use the formula \(q = mc\Delta T\). Here, \(m = 100.0\ g\), \(c = 4.184\ J/(g\cdot^{\circ}C)\), \(\Delta T=T_{final}-T_{initial}=29.3 - 21.0=8.3^{\circ}C\).

$$q_{water}=100.0\times4.184\times8.3$$

$$q_{water}=100.0\times4.184\times8.3 = 3472.72\ J$$

Step2: Calculate heat released by metal

Since \(q_{metal}=-q_{water}\) (heat lost by metal = heat gained by water in an isolated system), \(q_{metal}=- 3472.72\ J\). For the metal, \(m = 54.80\ g\), \(\Delta T=T_{final}-T_{initial}=29.3 - 100.0=-70.7^{\circ}C\). Use the formula \(q = mc\Delta T\) to solve for \(c\) (specific heat of metal). Rearranging for \(c\) gives \(c=\frac{q}{m\Delta T}\).

$$c=\frac{-3472.72}{54.80\times(- 70.7)}$$

$$c=\frac{-3472.72}{-3874.36}$$

$$c = 0.896\ J/(g\cdot^{\circ}C)$$

Answer:

The specific heat of the unknown metal sample is \(0.896\ J/(g\cdot^{\circ}C)\)