a 200 g aluminum sample is mixed with 200 g of water at 20.0°c in a calorimeter. the final equilibrium temperature of the mixture is 34.2°c. what was the initial temperature of the aluminum? refer to table 12.2.
100°c
120°c
160°c
140°c

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Accepted Answer

# Explanation:
## Step1: Recall Heat Transfer Formula
The heat lost by aluminum ($Q_{Al}$) equals heat gained by water ($Q_{w}$): $Q = mc\Delta T$, so $m_{Al}c_{Al}(T_{Al, initial}-T_{final}) = m_{w}c_{w}(T_{final}-T_{w, initial})$.

## Step2: Identify Constants
$m_{Al}=200\ g$, $m_{w}=200\ g$, $T_{w, initial}=20.0^\circ C$, $T_{final}=34.2^\circ C$, $c_{Al}=0.900\ J/g^\circ C$, $c_{w}=4.186\ J/g^\circ C$.

## Step3: Substitute Values
$$
\begin{align*}
200 \times 0.900 \times (T_{Al, initial}-34.2) &= 200 \times 4.186 \times (34.2 - 20.0)\
180 \times (T_{Al, initial}-34.2) &= 837.2 \times 14.2\
180 \times (T_{Al, initial}-34.2) &= 11888.24\
T_{Al, initial}-34.2 &= \frac{11888.24}{180}\
T_{Al, initial}-34.2 &\approx 66.05\
T_{Al, initial} &\approx 66.05 + 34.2\
T_{Al, initial} &\approx 100.25^\circ C \approx 100^\circ C
\end{align*}
$$

# Answer:
100°C (Option: 100°C)

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Question

Explanation:

Step1: Recall Heat Transfer Formula

Step2: Identify Constants

Step3: Substitute Values

Answer: