If a 136.51 g Al rod (c = 0.900 J/g ° C) is at 100.0 ° C and placed into 250.0 g of water at 21.8 ° C, what is the temperature of the system (rod + water) at equilibrium assuming no losses to the surroundings?

Question

Quizplus · Accepted Answer

To find the equilibrium temperature, we use the principle of conservation of energy, where the heat lost by the aluminum rod will equal the heat gained by the water. The formula for heat (q) is $q = mc\Delta T$, where $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is the change in temperature.For the aluminum rod, $q_{Al} = m_{Al}c_{Al}\Delta T_{Al}$, and for water, $q_{water} = m_{water}c_{water}\Delta T_{water}$.Given:- $m_{Al} = 136.51 g$,- $c_{Al} = 0.900 J/g°C$,- Initial $T_{Al} = 100.0°C$,- $m_{water} = 250.0 g$,- $c_{water} = 4.18 J/g°C$ (standard value for water),- Initial $T_{water} = 21.8°C$.Let $T_{final}$ be the final equilibrium temperature for both. Then,$\Delta T_{Al} = T_{final} - 100.0°C$,$\Delta T_{water} = T_{final} - 21.8°C$.Setting the heat lost by aluminum equal to the heat gained by water, we get:$$m_{Al}c_{Al}(T_{final} - 100.0) = m_{water}c_{water}(T_{final} - 21.8)$$.Solving for $T_{final}$ with the given values:$$136.51 \times 0.900 \times (T_{final} - 100.0) = 250.0 \times 4.18 \times (T_{final} - 21.8)$$.This equation simplifies to find $T_{final}$, which upon calculation, gives a value close to 30°C, making option C the correct answer.

If a 13651 G Al Rod (C = 0