Question

a piece of silver whose mass is 10.3 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:

Step1: Recall the heat - transfer formula

The formula for heat transfer is $Q = mc\Delta T$, where $Q$ is the heat energy absorbed or released, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. The specific heat capacity of water $c_{water}=4.18\ J/(g\cdot^{\circ}C)$ and the specific heat capacity of silver $c_{silver}=0.235\ J/(g\cdot^{\circ}C)$. The change in temperature $\Delta T=T_f - T_i=37.6 - 24.0=13.6^{\circ}C$.

Step2: Calculate the heat absorbed by water

For water, $m_{water}=20.9\ g$, $c_{water}=4.18\ J/(g\cdot^{\circ}C)$ and $\Delta T = 13.6^{\circ}C$. Using the formula $Q_{water}=m_{water}c_{water}\Delta T$, we have $Q_{water}=20.9\times4.18\times13.6$.
$Q_{water}=20.9\times4.18\times13.6 = 20.9\times56.848=1188.1232\approx1190\ J$.

Step3: Calculate the heat absorbed by silver

For silver, $m_{silver}=10.3\ g$, $c_{silver}=0.235\ J/(g\cdot^{\circ}C)$ and $\Delta T = 13.6^{\circ}C$. Using the formula $Q_{silver}=m_{silver}c_{silver}\Delta T$, we have $Q_{silver}=10.3\times0.235\times13.6$.
$Q_{silver}=10.3\times0.235\times13.6 = 10.3\times3.196=32.9188\approx32.9\ J$.

Answer:

The energy absorbed by water is approximately $1190\ J$ and the energy absorbed by silver is approximately $32.9\ J$.