- #1
ElPimiento
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Homework Statement
"A well-insulated bucket of negligible heat capacity contains 120 g of ice at 0°C. If 20 g of steam at 100°C is injected into the bucket, what is the final equilibrium temperature of the system?"
Homework Equations
$$Q_{fus} = m_{water}L_{fus}$$
$$Q_{vap} = m_{steam}L_{vap}$$
$$Q = mc_{water}\Delta T$$
Where ##L_{fus} = 334 \frac{kJ}{kg}##, ##L_{vap} = -2230 \frac{kJ}{kg}## (since the phase change is from gas to liquid), and ##c_{water} = 4.184 \frac{kJ}{kgK}##
The Attempt at a Solution
(I'm beginning to suspect the error is a computational one, or that I've given some term the wrong sign)
Since the container is well-insulated I can write that no heat is lost:
$$\begin{align*}
\sum Q & = 0 \\
Q_{fus} + m_1c_{water}\big(T_f - (273K)\big) + Q_{vap} + m_2c_{water}\big(T_f - (373K)\big) & = 0 \\
\end{align*}$$
Where ##m_1## refers to the ice and ##m_2## refers to the steam. Some algebra,
$$\begin{align*}
\big(m_1c_{water}T_f - m_1c_{water}(273K)\big) & \\
+ \big(m_2c_{water}T_f - m_2c_{water}(373K)\big) & = -(Q_{fus} + Q_{vap}) \\ \\
\big(m_1c_{water}T_f + m_2c_{water}T_f\big) & \\
- \big(m_1c_{water}(273K) + m_2c_{water}(373K)\big) & = -(Q_{fus} + Q_{vap}) \\ \\
T_fc_{water}(m_1 + m_2) & = -(Q_{fus} + Q_{vap}) \\
&\ \ \ \ + \big(m_1c_{water}(273K) + m_2c_{water}(373K)\big) \\ \\
T_f & = \frac {-(Q_{fus} + Q_{vap}) + \big(m_1c_{water}(273K) + m_2c_{water}(373K)\big)} {c_{water}(m_1 + m_2)}.
\end{align*}$$
Numerically, this becomes,
$$\begin{align*}
T_f & \approx \frac {-(-4.52 kJ) + \big((137 kJ) + (31.2 kJ)\big)} {0.586 \frac{kJ}{K}} \\
& \approx 294 K.
\end{align*}$$
or 21.7°C; but this is incorrect.
Thanks,
Andrew
