Question

An aluminum block of mass \(m_{\mathrm{Al}}=2.0 \mathrm{~kg}\) and specific heat \(c_{\mathrm{Al}}=910 \mathrm{~J} /(\mathrm{kg} \mathrm{K})\) is at an initial temperature of \(1000^{\circ} \mathrm{C}\) and is dropped into a bucket of water. The water has mass \(m_{\mathrm{H}_{2} \mathrm{O}}=12 \mathrm{~kg}\) and specific heat \(c_{\mathrm{H}_{2} \mathrm{O}}=4190 \mathrm{~J} /(\mathrm{kg} \mathrm{K})\) and is at room temperature \(\left(25^{\circ} \mathrm{C}\right) .\) What is the approximate final temperature of the system when it reaches thermal equilibrium? (Neglect heat loss out of the system.) a) \((\rho c / V) \Delta T\) b) \((\rho c V)(\Delta T+273.15)\) c) \((\rho c V) / \Delta T\) d) \(\rho c V \Delta T\)

Short answer

Answer: The final temperature of the system is approximately 1.014°C. The given answer choices do not align with the solution to the problem.

Step-by-step solution

Calculate the heat gained by the water (Q_water)

We'll use the formula \(Q = mc\Delta T\) to find the heat gained by the water, where \(m\) is the mass, \(c\) is the specific heat, and \(\Delta T\) is the change in temperature. Given \(m_{H_2O} = 12 \, kg\), \(c_{H_2O} = 4190 \, \frac{J}{kg \, K}\), and initial temperature \(T_{water} = 25^\circ C\), we get: \(Q_{water} = m_{H_2O} c_{H_2O} (T_f - T_{water})\), where \(T_f\) is the final temperature.

Calculate the heat lost by the aluminum block (Q_Al)

Likewise, we'll use the same formula for the aluminum block, given \(m_{Al} = 2.0 \, kg\), \(c_{Al} = 910 \, \frac{J}{kg \, K}\), and initial temperature \(T_{Al} = 1000^\circ C\): \(Q_{Al} = m_{Al} c_{Al} (T_{Al} - T_f)\).

Set up the equation and solve for the final temperature (T_f)

Since the heat gained by the water is equal to the heat lost by the aluminum block, we can set: \(Q_{water} = Q_{Al}\), thus, \(m_{H_2O} c_{H_2O} (T_f - T_{water}) = m_{Al} c_{Al} (T_{Al} - T_f)\). Now, we'll solve for \(T_f\): \(T_f(m_{H_2O} c_{H_2O} + m_{Al} c_{Al}) = m_{H_2O} c_{H_2O} T_{water} + m_{Al} c_{Al} T_{Al}\). Substituting the given values and solving for \(T_f\), we get: \(T_f(12 \cdot 4190 + 2 \cdot 910) = 12 \cdot 4190 \cdot 25 + 2 \cdot 910 \cdot 1000\), \(T_f \approx 54971.610 / 54180\), \(T_f \approx 1.014^\circ C\). The final temperature of the system is approximately \(1.014^\circ C\). It seems like the given answer choices can't be matched with the solution found, meaning that the possible answers provided don't align with the solution to the problem.