Question

a 37.3 g sample of copper at 99.8 °c is carefully placed into an insulated container containing 205 g of water at 18.5 °c. calculate the final temperature when thermal equilibrium is reached. assume there is no energy transferred to or from the container. specific heat capacities: cu = 0.385 j g⁻¹ °c⁻¹, h₂o = 4.184 j g⁻¹ °c⁻¹

Explanation:

Step1: Set up heat - transfer equation

According to the principle of conservation of energy, the heat lost by copper is equal to the heat gained by water. The heat - transfer formula is $Q = mc\Delta T$. Let the final temperature be $T$. The heat lost by copper $Q_{Cu}=m_{Cu}c_{Cu}(T_{Cu}-T)$, and the heat gained by water $Q_{H_2O}=m_{H_2O}c_{H_2O}(T - T_{H_2O})$. So $m_{Cu}c_{Cu}(T_{Cu}-T)=m_{H_2O}c_{H_2O}(T - T_{H_2O})$.

Step2: Substitute given values

We know that $m_{Cu}=37.3\ g$, $c_{Cu}=0.385\ Jg^{-1}{^{\circ}C}^{-1}$, $T_{Cu}=99.8^{\circ}C$, $m_{H_2O}=205\ g$, $c_{H_2O}=4.184\ Jg^{-1}{^{\circ}C}^{-1}$, $T_{H_2O}=18.5^{\circ}C$.
Substituting these values into the equation:

$$37.3\times0.385\times(99.8 - T)=205\times4.184\times(T - 18.5)$$

First, calculate the left - hand side: $37.3\times0.385 = 14.3605$, so the left - hand side is $14.3605\times(99.8 - T)=14.3605\times99.8-14.3605T=1433.1779 - 14.3605T$.
Then, calculate the right - hand side: $205\times4.184 = 857.72$, so the right - hand side is $857.72\times(T - 18.5)=857.72T-857.72\times18.5 = 857.72T - 15867.82$.

Step3: Solve for $T$

$$1433.1779 - 14.3605T=857.72T-15867.82$$

Add $14.3605T$ to both sides:

$$1433.1779=857.72T + 14.3605T-15867.82$$

$$1433.1779=872.0805T-15867.82$$

Add $15867.82$ to both sides:

$$1433.1779 + 15867.82=872.0805T$$

$$17300.9979 = 872.0805T$$

$$T=\frac{17300.9979}{872.0805}\approx19.8^{\circ}C$$

Answer:

$19.8^{\circ}C$