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The concept of the "heat of fusion" is fundamental in understanding how substances change from solid to liquid. Specifically for water, the heat of fusion refers to the energy needed to change 1 mole of ice at its melting point, without a change in temperature, into liquid water.
The heat of fusion is crucial in this exercise.
This is because when the ice melts, it absorbs heat from its surroundings.

• For water, the heat of fusion is given as 6.01 kJ/mol.
• This means that to melt 1 mole of ice, 6010 J of energy is absorbed.

The formula used in the exercise to determine the mass of ice is \( \Delta Q_{\text{ice}} = n_{\text{ice}} \cdot \Delta H_{f} \).
Here, \( \Delta Q_{\text{ice}} \) is the heat absorbed by the ice, \( n_{\text{ice}} \) is the number of moles, and \( \Delta H_{f} \) is the heat of fusion.
This relationship allows us to calculate how much ice melts by relating the energy absorbed to the energy supplied.

The "specific heat capacity" is another essential concept. It measures how much heat a substance must absorb to increase its temperature by 1 degree Celsius.
In this exercise, we look at the specific heat capacity of water which is 4.18 J/(g°C).

• It tells us how water requires 4.18 joules to raise the temperature of 1 gram of water by 1°C.
• This value is crucial for knowing how much heat will be lost or gained during temperature changes.

The exercise uses the specific heat capacity formula to figure out how much heat the water loses when it cools down from 21°C to 0°C:\[\Delta Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}}\]In this formula, \( m_{\text{water}} \) is the mass of the water, \( c_{\text{water}} \) is the specific heat, and \( \Delta T_{\text{water}} \) is the temperature difference.
This calculation is key because the heat lost by the water equals the heat absorbed by the ice during melting.

The concept of "temperature change" is pivotal when exploring heat exchanges.
It's about understanding the shift in thermal energy as substances (like water in this case) move to different temperatures.

• For this problem, the water's temperature change is from 21°C down to 0°C, which involves a decrease.
• The temperature difference, or \( \Delta T \), here is simply 21°C.

This change is critical since cooling the water from 21°C to 0°C means heat must be released from the water.
This released heat is then absorbed by the ice to allow it to melt and reach 0°C.
Determining how much heat is involved with the formula \( \Delta Q = m \cdot c \cdot \Delta T \) is what links the concepts together into a complete understanding of the exercise's solution.