Solution to Question 8
We have two calorimeters A and B with equal amounts of water ($300 \text{ g}$ each).
- Initial Temperature of A ($T_A$) = $48^\circ\text{C}$
- Initial Temperature of B ($T_B$) = $80^\circ\text{C}$
- Initial difference $\Delta T = 80 – 48 = 32^\circ\text{C}$
Step 1 (A $\to$ B): Transfer $100 \text{ g}$ from A ($48^\circ\text{C}$) to B ($300 \text{ g}$, $80^\circ\text{C}$). B will now have $400 \text{ g}$. The new temperature of B ($T’_B$) is the weighted average:
$$ T’_B = \frac{1(48) + 3(80)}{4} = \frac{288}{4} = 72^\circ\text{C} $$
A has $200 \text{ g}$ at $48^\circ\text{C}$. B has $400 \text{ g}$ at $72^\circ\text{C}$.
Step 2 (B $\to$ A): Transfer $100 \text{ g}$ from B ($72^\circ\text{C}$) back to A ($200 \text{ g}$, $48^\circ\text{C}$). A returns to $300 \text{ g}$. The new temperature of A ($T’_A$) is:
$$ T’_A = \frac{1(72) + 2(48)}{3} = \frac{168}{3} = 56^\circ\text{C} $$
Result after Cycle 1: $T_A = 56^\circ\text{C}$, $T_B = 72^\circ\text{C}$.
New Difference $\Delta T_1 = 72 – 56 = 16^\circ\text{C}$.
Notice that the difference reduced from $32$ to $16$. It halved.
Since the masses and transfer amounts are constant, the reduction factor for the temperature difference is constant per cycle. In each complete cycle, the temperature difference is reduced by half.
Sequence of temperature differences ($\Delta T$):
- Start: $32^\circ\text{C}$
- Cycle 1: $16^\circ\text{C}$
- Cycle 2: $8^\circ\text{C}$
- Cycle 3: $4^\circ\text{C}$
- Cycle 4: $2^\circ\text{C}$
- Cycle 5: $1^\circ\text{C}$
The question asks how many times the cycle must be repeated until the difference becomes less than $1.0^\circ\text{C}$.
At Cycle 5, the difference is exactly $1.0^\circ\text{C}$. To be strictly less than $1.0^\circ\text{C}$, one might argue for the 6th cycle.
Answer: 5 times