Question

How much energy is required to change a 46 g ice cube from ice at −10◦C to steam at 112◦C? The specific heat of ice is 2090 J/kg ·◦ C, the specific heat of water is 4186 J/kg ·◦ C, the specific heat of stream is 2010 J/kg ·◦ C, the heat of fusion is 3.33 × 105 J/kg, and the heat of vaporization is 2.26 × 106 J/kg. Answer in units of J.

Answer 1

To calculate the total energy required to change a 46 g ice cube from ice at -10°C to steam at 112°C, we need to break the process down into the following steps:

1. Heat the ice from -10°C to 0°C.
2. Melt the ice at 0°C to water.
3. Heat the water from 0°C to 100°C.
4. Vaporize the water at 100°C to steam.
5. Heat the steam from 100°C to 112°C.

### Step 1: Heat the ice from -10°C to 0°C

We use the formula:
\[
Q = m \cdot c \cdot \Delta T
\]
where:
- \( m = 0.046 \, \text{kg} \) (mass of the ice)
- \( c = 2090 \, \text{J/kg} \cdot {}^\circ C \) (specific heat of ice)
- \( \Delta T = 0 - (-10) = 10^\circ C \)

So,
\[
Q_1 = 0.046 \, \text{kg} \cdot 2090 \, \text{J/kg} \cdot {}^\circ C \cdot 10^\circ C = 0.046 \cdot 2090 \cdot 10
\]
\[
Q_1 = 0.046 \cdot 20900 = 962.3 \, \text{J}
\]

### Step 2: Melt the ice at 0°C to water

We use the formula:
\[
Q = m \cdot L_f
\]
where \( L_f = 3.33 \times 10^5 \, \text{J/kg} \).

So,
\[
Q_2 = 0.046 \, \text{kg} \cdot 3.33 \times 10^5 \, \text{J/kg} = 0.046 \cdot 333000
\]
\[
Q_2 = 15318 \, \text{J}
\]

### Step 3: Heat the water from 0°C to 100°C

Again using the formula:
\[
Q = m \cdot c \cdot \Delta T
\]
where:
- \( c = 4186 \, \text{J/kg} \cdot {}^\circ C \) (specific heat of water)
- \( \Delta T = 100 - 0 = 100^\circ C \)

So,
\[
Q_3 = 0.046 \, \text{kg} \cdot 4186 \, \text{J/kg} \cdot {}^\circ C \cdot 100^\circ C = 0.046 \cdot 418600
\]
\[
Q_3 = 19255.6 \, \text{J}
\]

### Step 4: Vaporize the water at 100°C to steam

Using the formula:
\[
Q = m \cdot L_v
\]
where \( L_v = 2.26 \times 10^6 \, \text{J/kg} \).

So,
\[
Q_4 = 0.046 \, \text{kg} \cdot 2.26 \times 10^6 \, \text{J/kg} = 0.046 \cdot 2260000
\]
\[
Q_4 = 10316 \, \text{J}
\]

### Step 5: Heat the steam from 100°C to 112°C

Using the formula:
\[
Q = m \cdot c \cdot \Delta T
\]
where:
- \( c = 2010 \, \text{J/kg} \cdot {}^\circ C \) (specific heat of steam)
- \( \Delta T = 112 - 100 = 12^\circ C \)

So,
\[
Q_5 = 0.046 \, \text{kg} \cdot 2010 \, \text{J/kg} \cdot {}^\circ C \cdot 12^\circ C = 0.046 \cdot 2010 \cdot 12
\]
\[
Q_5 = 1111.44 \, \text{J}
\]

### Total Energy Required

Now we add all the energies calculated:
\[
Q_{\text{total}} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5
\]
\[
Q_{\text{total}} = 962.3 + 15318 + 19255.6 + 10316 + 1111.44
\]
\[
Q_{\text{total}} \approx 36763.34 \, \text{J}
\]

Thus, the total energy required to change the 46 g ice cube from ice at -10°C to steam at 112°C is approximately:

\[
\boxed{36763 \, \text{J}}
\]