Sure! Let's walk through the problem step by step to find the final temperature of the water when a 1000 kW heater supplies heat to it.
1. Understanding the Problem:
- We are given:
- The power of the heater: [tex]\(1000\)[/tex] kW.
- The heat supplied: [tex]\(4.2 \times 10^5\)[/tex] kJ.
- The mass of the water: [tex]\(20\)[/tex] kg.
- The initial temperature of the water: [tex]\(30^{\circ}C\)[/tex].
- The specific heat capacity of water: [tex]\(4.186\)[/tex] kJ/(kg·°C).
2. Calculating the Heat Supplied:
- The heat supplied in kilojoules is given directly as [tex]\(4.2 \times 10^5\)[/tex] kJ.
3. Calculating the Temperature Change of Water:
- To determine the change in temperature ([tex]\(\Delta T\)[/tex]), we use the formula:
[tex]\[
Q = m \cdot C \cdot \Delta T
\][/tex]
where:
- [tex]\(Q\)[/tex] is the heat supplied (in kJ),
- [tex]\(m\)[/tex] is the mass of water (in kg),
- [tex]\(C\)[/tex] is the specific heat capacity of water (in kJ/(kg·°C)),
- [tex]\(\Delta T\)[/tex] is the change in temperature (in °C).
Rearranging the formula to solve for [tex]\(\Delta T\)[/tex], we get:
[tex]\[
\Delta T = \frac{Q}{m \cdot C}
\][/tex]
Substituting the given values into the equation:
[tex]\[
\Delta T = \frac{4.2 \times 10^5 \text{kJ}}{20 \text{ kg} \times 4.186 \text{ kJ/(kg·°C)}}
\][/tex]
4. Calculating the Numerical Value of [tex]\(\Delta T\)[/tex]:
- Performing the division:
[tex]\[
\Delta T \approx 5016.722 \, ^{\circ}\text{C}
\][/tex]
5. Determining the Final Temperature:
- The final temperature ([tex]\(T_{\text{final}}\)[/tex]) is found by adding the change in temperature ([tex]\(\Delta T\)[/tex]) to the initial temperature ([tex]\(T_{\text{initial}}\)[/tex]):
[tex]\[
T_{\text{final}} = T_{\text{initial}} + \Delta T
\][/tex]
- Substituting the initial temperature and [tex]\(\Delta T\)[/tex]:
[tex]\[
T_{\text{final}} = 30^{\circ}\text{C} + 5016.722^{\circ}\text{C}
\][/tex]
[tex]\[
T_{\text{final}} \approx 5046.722 \, ^{\circ}\text{C}
\][/tex]
Therefore, the final temperature of the water, after the 1000 kW heater supplies [tex]\(4.2 \times 10^5\)[/tex] kJ of heat to [tex]\(20\)[/tex] kg of water initially at [tex]\(30^{\circ}\text{C}\)[/tex], is approximately [tex]\(5046.722 \, ^{\circ}\text{C}\)[/tex].
