Solution Found!
A 15.7-g aluminum block is warmed to 53.2 °C and plunged into an insulated beaker
Chapter 3, Problem 107P(choose chapter or problem)
A 15.7-g aluminum block is warmed to \(53.2\ ^{\circ}\mathrm{C}\) and plunged into an insulated beaker containing 32.5 g of water initially at \(24.5\ ^{\circ}\mathrm{C}\). The aluminum and the water are allowed to come to thermal equilibrium. Assuming that no heat is lost, what is the final temperature of the water and aluminum?
Questions & Answers
QUESTION:
A 15.7-g aluminum block is warmed to \(53.2\ ^{\circ}\mathrm{C}\) and plunged into an insulated beaker containing 32.5 g of water initially at \(24.5\ ^{\circ}\mathrm{C}\). The aluminum and the water are allowed to come to thermal equilibrium. Assuming that no heat is lost, what is the final temperature of the water and aluminum?
ANSWER:Step 1 of 3
Here, we are going to calculate the final temperature of the water and Al.
When two systems are combined together, thermal energy is transferred from the hotter body to the cooler one. Thus, the heat absorbed by one system is equal to the heat released by another system
Therefore,
\(\mathrm{Q}_{\text {system (1) }}=-\mathrm{Q}_{\text {system (2) }}\)
Thermal energy transfers from metal to water.
From equation(1),
\(\begin{array}{l} \mathrm{Q}_{\text {metal }}=-\mathrm{Q}_{\text {water }} \\ \Rightarrow \mathrm{m}_{\text {metal }} \times \mathrm{C}_{\text {s metal }} \times \Delta T_{\text {metal }}=-\mathrm{m}_{\text {water }} \times \mathrm{C}_{\text {s water }} \times \Delta T_{\text {water }} \end{array}\)