Question

a copper slug of mass m = 75.0 g is heated in a laboratory oven to a temperature of t = 312 c. the slug is then dropped into a glass beaker of mass $m_{g}=50.0$ g containing a mass $m_{w}=220.0$ g of water. the initial temperature of the water and the beaker is $t_{i}=12.0$ c. assuming that the slug, beaker and water are an isolated system, and the water does not vaporize, find the final temperature of the system at thermal equilibrium. (heat capacities of water $c_{w}=4187$, copper $c_{c}=386$, glass $c_{g}=840$ j/kgk) copper $t_{h}=312^{circ}c$ $varphi(hot)=mcc(t_{f}-t_{h})$ $varphi(w)=m_{w}c_{w}(t_{f}-t_{c})$ $varphi(g)=m_{g}c_{g}(t_{f}-t_{c})$ $varphi(hot)+varphi(cold)=0$ $varphi(hot)+varphi(w)+varphi(glass)=0$ $varphi(w)+varphi(glass)=-varphi(hot)$ $mc(t_{f}-t_{h})+m_{w}c_{w}(t_{f}-t_{c})+m_{g}c_{g}(t_{f}-t_{c}) = 0$ $t_{f}(mc + m_{w}c_{w}+m_{g}c_{g})=mct_{h}+t_{c}(m_{w}c_{w}+m_{g}c_{g})$ $t_{f}=\frac{mct_{h}+t_{c}(m_{w}c_{w}+m_{g}c_{g})}{(mc + m_{w}c_{w}+m_{g}c_{g})}$ $t_{f}=\frac{(0.075\times386\times312)+12.0(0.220\times4187 + 0.050\times840)}{(0.075\times386)+(0.220\times4187)+(0.050\times840)}$

Explanation:

Step1: Identify the heat - transfer principle

In an isolated system, the sum of heat lost by the hot object and heat gained by the cold objects is zero, i.e., $\sum Q = 0$. Here, the hot object is the copper slug and the cold objects are the water and the glass beaker. 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.

Step2: Write the heat - balance equation

Let the final temperature of the system be $T_f$. The heat lost by the copper slug is $Q_{copper}=m_{c}c_{c}(T_{h}-T_{f})$, the heat gained by the water is $Q_{water}=m_{w}c_{w}(T_{f}-T_{c})$, and the heat gained by the glass beaker is $Q_{glass}=m_{g}c_{g}(T_{f}-T_{c})$. So, $m_{c}c_{c}(T_{h}-T_{f})+m_{w}c_{w}(T_{f}-T_{c})+m_{g}c_{g}(T_{f}-T_{c}) = 0$.

Step3: Rearrange the equation for $T_f$

Expanding the above equation gives $m_{c}c_{c}T_{h}-m_{c}c_{c}T_{f}+m_{w}c_{w}T_{f}-m_{w}c_{w}T_{c}+m_{g}c_{g}T_{f}-m_{g}c_{g}T_{c}=0$. Grouping the terms with $T_f$ on one side: $T_{f}(m_{c}c_{c}+m_{w}c_{w}+m_{g}c_{g})=m_{c}c_{c}T_{h}+T_{c}(m_{w}c_{w}+m_{g}c_{g})$. Then $T_{f}=\frac{m_{c}c_{c}T_{h}+T_{c}(m_{w}c_{w}+m_{g}c_{g})}{m_{c}c_{c}+m_{w}c_{w}+m_{g}c_{g}}$.

Step4: Substitute the given values

We have $m_{c}=75.0\ g = 0.075\ kg$, $c_{c}=386\ J/(kg\cdot K)$, $T_{h}=312^{\circ}C$, $m_{w}=220.0\ g = 0.220\ kg$, $c_{w}=4187\ J/(kg\cdot K)$, $m_{g}=50.0\ g = 0.050\ kg$, $c_{g}=840\ J/(kg\cdot K)$, and $T_{c}=12.0^{\circ}C$.
\[

$$\begin{align*} T_{f}&=\frac{0.075\times386\times312 + 12\times(0.220\times4187+0.050\times840)}{0.075\times386+(0.220\times4187)+(0.050\times840)}\\ &=\frac{0.075\times386\times312+12\times(921.14 + 42)}{28.95+921.14 + 42}\\ &=\frac{8878.2+12\times963.14}{992.09}\\ &=\frac{8878.2 + 11557.68}{992.09}\\ &=\frac{20435.88}{992.09}\\ &\approx20.6^{\circ}C \end{align*}$$

\]

Answer:

$20.6^{\circ}C$