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) \(50^{\circ} \mathrm{C}\) b) \(60^{\circ} \mathrm{C}\) c) \(70^{\circ} \mathrm{C}\) d) \(80^{\circ} \mathrm{C}\)

Step by step solution

01

Calculate the energy lost by the aluminum block

To calculate the energy lost by the aluminum block, we can use the formula: \(q_{Al} = M_{Al} \times C_{Al} \times (T_{initial} - T_{final})\) where \(T_{initial}\) is the initial temperature of the aluminum block (1000 °C), and \(T_{final}\) is the final temperature we want to determine. We will rewrite the formula above, leaving \(T_{final}\) as the subject: \(T_{final} = T_{initial} - \frac{q_{Al}}{M_{Al} \times C_{Al}}\)

02

Calculate the energy gained by the water

To calculate the energy gained by the water, we can use the formula: \(q_{H_2O} = M_{H_2O} \times C_{H_2O} \times (T_{final} - T_{initial})\) where \(T_{initial}\) is the initial temperature of the water (25 °C), and \(T_{final}\) is the final temperature we want to determine. We will rewrite the formula above, leaving \(q_{H_2O}\) as the subject: \(q_{H_2O} = M_{H_2O} \times C_{H_2O} \times (T_{final} - T_{initial})\)

03

Apply conservation of energy

As stated before, the energy lost by the aluminum block must be equal to the energy gained by the water. Therefore, we can equate the two formulas from Step 1 and Step 2: \(M_{Al} \times C_{Al} \times (T_{initial} - T_{final}) = M_{H_2O} \times C_{H_2O} \times (T_{final} - T_{initial})\) Now we can plug in the given values for mass and specific heat and solve for \(T_{final}\): \(2.0\ Kg \times 910\ J/(KgK) \times (1000 - T_{final}) = 12.0\ Kg \times 4190\ J/(KgK) \times (T_{final} - 25)\)

04

Solve for the final temperature

To solve for \(T_{final}\), first expand the equation and simplify: \(1820(1000 - T_{final}) = 50280(T_{final} - 25)\) \(1820000 - 1820T_{final} = 50280T_{final} - 1257000\) Now combine like terms and solve for \(T_{final}\): \(1820T_{final} + 50280T_{final} = 1820000 - 1257000\) \(52100T_{final} = 563000\) \(T_{final} = 563000 \div 52100\) \(T_{final} \approx 10.8\) Since the initial temperature of the water was 25 °C, the final temperature is about 25 + 10.8 = 35.8 °C, which is closest to the given option a) \(50^{\circ} \mathrm{C}.\) Thus, the approximate final temperature of the system when it reaches thermal equilibrium is 50 °C.

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