A 0.0500 kg ice cube starts at -18.5°C. How much heat does it take to melt it completely? (Note: Warm it up to the melting point first.)

| Material | Melt Pt (°C) | L (J/kg) | Boil Pt (°C) | L (J/kg) | C (J/(kg·°C)) |
|----------|--------------|----------|--------------|----------|---------------|
| Water | 0 | 3.33 × 10^5 | 100 | 2.26 × 10^6 | 4186 |

(Unit = J)



Answer :

To determine the total amount of heat required to melt a [tex]$0.0500 \, \text{kg}$[/tex] (50 grams) ice cube starting at [tex]$-18.5^{\circ} \text{C}$[/tex], we need to perform the following steps:

1. Calculate the amount of heat required to raise the temperature of the ice from [tex]$-18.5^{\circ} \text{C}$[/tex] to [tex]$0^{\circ} \text{C}$[/tex].

2. Calculate the amount of heat required to melt the ice at [tex]$0^{\circ} \text{C}$[/tex] into water.

3. Sum the heats calculated in the previous steps to find the total heat required.

### Step 1: Heating the Ice to [tex]$0^{\circ} \text{C}$[/tex]

The formula to calculate the heat required to change the temperature of a substance is:

[tex]\[ q = m \cdot c \cdot \Delta T \][/tex]

Where:
- [tex]\( q \)[/tex] is the heat energy (in joules, [tex]\( J \)[/tex]).
- [tex]\( m \)[/tex] is the mass of the substance (in kilograms, [tex]\( kg \)[/tex]).
- [tex]\( c \)[/tex] is the specific heat capacity (in joules per kilogram per degree Celsius, [tex]\( J/(kg \cdot ^{\circ}C) \)[/tex]).
- [tex]\( \Delta T \)[/tex] is the change in temperature (final temperature - initial temperature).

For ice, the specific heat capacity [tex]\( c_{\text{ice}} = 2090 \, J/(kg \cdot ^{\circ}C) \)[/tex].

So, we need to find the heat required to raise the temperature of [tex]$0.0500 \, \text{kg}$[/tex] of ice from [tex]$-18.5^{\circ} \text{C}$[/tex] to [tex]$0^{\circ} \text{C}$[/tex]:

[tex]\[ \Delta T = 0^{\circ}C - (-18.5^{\circ}C) = 18.5^{\circ}C \][/tex]

[tex]\[ q_{\text{heat ice}} = 0.0500 \, \text{kg} \cdot 2090 \, J/(kg \cdot ^{\circ}C) \cdot 18.5^{\circ}C \][/tex]

[tex]\[ q_{\text{heat ice}} = 1933.25 \, J \][/tex]

### Step 2: Melting the Ice at [tex]$0^{\circ} \text{C}$[/tex]

The heat required to melt ice is determined by the latent heat of fusion [tex]\( L_{\text{fusion}} \)[/tex], which is the amount of heat required to change [tex]$1 \, \text{kg}$[/tex] of a solid into a liquid without a temperature change.

For water, [tex]\( L_{\text{fusion}} = 3.33 \times 10^5 \, J/kg \)[/tex].

The heat required to melt [tex]$0.0500 \, \text{kg}$[/tex] of ice is:

[tex]\[ q_{\text{melt ice}} = m \cdot L_{\text{fusion}} \][/tex]

[tex]\[ q_{\text{melt ice}} = 0.0500 \, \text{kg} \cdot 3.33 \times 10^5 \, J/kg \][/tex]

[tex]\[ q_{\text{melt ice}} = 16650.0 \, J \][/tex]

### Step 3: Total Heat Required

Finally, sum up the heat required to warm the ice to [tex]$0^{\circ} \text{C}$[/tex] and the heat required to melt it:

[tex]\[ q_{\text{total}} = q_{\text{heat ice}} + q_{\text{melt ice}} \][/tex]

[tex]\[ q_{\text{total}} = 1933.25 \, J + 16650.0 \, J \][/tex]

[tex]\[ q_{\text{total}} = 18583.25 \, J \][/tex]

Thus, the total amount of heat required to melt the [tex]$0.0500 \, \text{kg}$[/tex] ice cube completely, starting from [tex]$-18.5^{\circ} \text{C}$[/tex], is 18583.25 Joules.

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