Question

1. a silver block, initially at 58.5°c, is submerged into 100.0 g of water at 24.8°c, in an insulated container. the final temperature of the mixture upon reaching thermal equilibrium is 26.2°c. what is the mass of the silver block?

Explanation:

Step1: Recall heat - transfer formula

The heat lost by the silver block ($q_{Ag}$) is equal to the heat gained by the water ($q_{w}$) since the system is insulated ($q_{Ag}=-q_{w}$). The heat - transfer formula is $q = mc\Delta T$, where $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. The specific heat capacity of water $c_{w}=4.184\ J/(g\cdot^{\circ}C)$ and the specific heat capacity of silver $c_{Ag}=0.235\ J/(g\cdot^{\circ}C)$.

Step2: Express $q_{Ag}$ and $q_{w}$

$q_{Ag}=m_{Ag}c_{Ag}(T_{f}-T_{i,Ag})$ and $q_{w}=m_{w}c_{w}(T_{f}-T_{i,w})$.
Since $q_{Ag}=-q_{w}$, we have $m_{Ag}c_{Ag}(T_{f}-T_{i,Ag})=-m_{w}c_{w}(T_{f}-T_{i,w})$.

Step3: Substitute the known values

We know that $m_{w} = 100.0\ g$, $c_{w}=4.184\ J/(g\cdot^{\circ}C)$, $c_{Ag}=0.235\ J/(g\cdot^{\circ}C)$, $T_{i,Ag}=58.5^{\circ}C$, $T_{i,w}=24.8^{\circ}C$, and $T_{f}=26.2^{\circ}C$.
$m_{Ag}\times0.235\ J/(g\cdot^{\circ}C)\times(26.2 - 58.5)^{\circ}C=- 100.0\ g\times4.184\ J/(g\cdot^{\circ}C)\times(26.2 - 24.8)^{\circ}C$.

Step4: Solve for $m_{Ag}$

First, calculate the right - hand side:
$-100.0\ g\times4.184\ J/(g\cdot^{\circ}C)\times(26.2 - 24.8)^{\circ}C=-100.0\times4.184\times1.4\ J=-585.76\ J$.
Then, calculate the left - hand side:
$m_{Ag}\times0.235\ J/(g\cdot^{\circ}C)\times(26.2 - 58.5)^{\circ}C=m_{Ag}\times0.235\ J/(g\cdot^{\circ}C)\times(-32.3)^{\circ}C=-7.5905m_{Ag}\ J/g$.
Set the two sides equal:
$-7.5905m_{Ag}=-585.76$.
$m_{Ag}=\frac{585.76}{7.5905}\ g\approx77.2\ g$.

Answer:

$77.2\ g$