Question

a piece of silver whose mass is 10.4 g is immersed in 20.9 g of water. the system is heated electrically from 24.0 to 37.6°c. how many joules of energy are absorbed by the silver? how many joules of energy are absorbed by the water?

Explanation:

For energy absorbed by silver:

Step1: Recall the formula for heat absorption

The formula for heat absorbed \( q \) is \( q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature. For silver, \( c_{\text{Ag}} = 0.235 \, \text{J/g°C} \), \( m = 10.4 \, \text{g} \), and \( \Delta T = 37.6 - 24.0 = 13.6 \, \text{°C} \).

Step2: Substitute values into the formula

\( q_{\text{Ag}} = 10.4 \, \text{g} \times 0.235 \, \text{J/g°C} \times 13.6 \, \text{°C} \)
First, calculate \( 10.4 \times 0.235 = 2.444 \)
Then, \( 2.444 \times 13.6 \approx 33.2 \, \text{J} \)

Step1: Recall the formula for heat absorption

Use \( q = mc\Delta T \). For water, \( c_{\text{H}_2\text{O}} = 4.184 \, \text{J/g°C} \), \( m = 20.9 \, \text{g} \), and \( \Delta T = 13.6 \, \text{°C} \) (same as above).

Step2: Substitute values into the formula

\( q_{\text{H}_2\text{O}} = 20.9 \, \text{g} \times 4.184 \, \text{J/g°C} \times 13.6 \, \text{°C} \)
First, \( 20.9 \times 4.184 \approx 87.4456 \)
Then, \( 87.4456 \times 13.6 \approx 1189 \, \text{J} \) (the slight difference might be due to rounding during steps, but the calculation is consistent)

Answer:

\( 33.2 \)

For energy absorbed by water: