- #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