To solve the first question about turning water into steam, we need to calculate the heat required for the phase change from liquid water to steam. The process of turning water into steam at the boiling point (100 °C) is called vaporization, and it requires a specific amount of heat called the latent heat of vaporization.
The formula to calculate the heat required for the phase change is:
\[ Q = m \cdot L_v \]
where:
- \( Q \) is the heat required (in kJ)
- \( m \) is the mass of the substance (in kg)
- \( L_v \) is the latent heat of vaporization (in kJ/kg)
Given that:
- The mass of water, \( m \), is 12 kg
- The latent heat of vaporization of water, \( L_v \), is 2260 kJ/kg
Plugging in the given values:
\[ Q = 12 \, \text{kg} \times 2260 \, \text{kJ/kg} \]
\[ Q = 27120 \, \text{kJ} \]
So, the heat required to turn 12 kg of water into steam at 100 °C is 27120 kJ.
For the second question, to find out if 0.5 kg of ice at 0 °C melts, we need to calculate the heat required to melt the ice and compare it to the available heat. The process of turning ice into water at 0 °C is called fusion, and it requires a specific amount of heat called the latent heat of fusion.
The formula to calculate the heat required to melt the ice is:
\[ Q = m \cdot L_f \]
where:
- \( Q \) is the heat required (in Joules)
- \( m \) is the mass of the substance (in kg)
- \( L_f \) is the latent heat of fusion (in J/kg)
Given that:
- The mass of ice, \( m \), is 0.5 kg
- The latent heat of fusion of ice, \( L_f \), is 334000 J/kg
- The heat available is 325000 Joules
Let's calculate the heat required to melt the ice:
\[ Q = 0.5 \, \text{kg} \times 334000 \, \text{J/kg} \]
\[ Q = 167000 \, \text{J} \]
We compare the heat required with the available heat:
- Heat required to melt the ice: 167000 Joules
- Heat available: 325000 Joules
Since the heat available (325000 Joules) is greater than the heat required to melt the ice (167000 Joules), it is sufficient to melt the 0.5 kg of ice.
Thus, the 0.5 kg of ice at 0 °C does melt with 325000 Joules of heat available.
