Question

question 12 (1 point)
a 0.10-kg piece of copper, initially at 95°c, is dropped into 0.20 kg of water contained in a 0.28-kg aluminum can; the water and aluminum are initially at 15°c. what is the final temperature of the system?
19.2°c
18.3°c
17.8°c
23.7°c

Explanation:

Step1: Recall heat transfer formula

The heat lost by the copper (\(Q_{lost}\)) is equal to the heat gained by the water (\(Q_{water}\)) and the heat gained by the aluminum can (\(Q_{aluminum}\)) at thermal equilibrium. The formula for heat transfer is \(Q = mc\Delta T\), where \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T\) is the change in temperature.

Specific heat capacities (from standard values):

• \(c_{copper} = 387 \, \text{J/(kg·°C)}\)
• \(c_{water} = 4186 \, \text{J/(kg·°C)}\)
• \(c_{aluminum} = 900 \, \text{J/(kg·°C)}\)

Let the final temperature be \(T_f\).

Heat lost by copper: \(Q_{copper} = m_{copper}c_{copper}(T_{copper, initial} - T_f)\)
Heat gained by water: \(Q_{water} = m_{water}c_{water}(T_f - T_{water, initial})\)
Heat gained by aluminum: \(Q_{aluminum} = m_{aluminum}c_{aluminum}(T_f - T_{aluminum, initial})\)

At equilibrium, \(Q_{copper} = Q_{water} + Q_{aluminum}\)

Step2: Substitute values

\(m_{copper} = 0.10 \, \text{kg}\), \(T_{copper, initial} = 95^\circ\text{C}\)
\(m_{water} = 0.20 \, \text{kg}\), \(T_{water, initial} = 15^\circ\text{C}\)
\(m_{aluminum} = 0.28 \, \text{kg}\), \(T_{aluminum, initial} = 15^\circ\text{C}\)

Substitute into the equation:

$$ 0.10 \times 387 \times (95 - T_f) = 0.20 \times 4186 \times (T_f - 15) + 0.28 \times 900 \times (T_f - 15) $$

Step3: Simplify left and right sides

Left side: \(38.7 \times (95 - T_f) = 3676.5 - 38.7T_f\)

Right side: First, factor out \((T_f - 15)\):

$$ (T_f - 15)(0.20 \times 4186 + 0.28 \times 900) $$

Calculate the coefficients:
\(0.20 \times 4186 = 837.2\)
\(0.28 \times 900 = 252\)
Sum: \(837.2 + 252 = 1089.2\)
So right side: \(1089.2(T_f - 15) = 1089.2T_f - 16338\)

Step4: Solve for \(T_f\)

Set left = right:

$$ 3676.5 - 38.7T_f = 1089.2T_f - 16338 $$

Bring all \(T_f\) terms to right and constants to left:

$$ 3676.5 + 16338 = 1089.2T_f + 38.7T_f $$

$$ 20014.5 = 1127.9T_f $$

$$ T_f = \frac{20014.5}{1127.9} \approx 17.75^\circ\text{C} \approx 17.8^\circ\text{C} $$

Answer:

\(17.8^\circ\text{C}\) (corresponding to the option \(17.8^\circ\text{C}\))