Answer:
207.33 kJ.
Step-by-step explanation:
What is given?
Mass of water (m) = 400 g.
ΔT (Change of temperature) = 114 °C - ( - 10 °C) = 124 °C.
Specific heat of water (c) = 4.18 J/g°C.
Step-by-step solution:
To solve this problem we have to use the following formula:
![Q=m\cdot c\cdot\Delta T.](https://img.qammunity.org/2023/formulas/chemistry/college/71yqvegcd8do9ngybblmpvz87uqbmx27w3.png)
Where Q is heat, m is the mass, c is the specific heat of the water and ΔT is the change of temperature ( T final - T initial).
We want to find the value of Q, so we replace the data that we have:
![\begin{gathered} Q=400\text{ g}\cdot418\frac{J}{g\text{ \degree C}}\cdot124\text{ \degree C,} \\ Q=207328\text{ J.} \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/5du2ntvfgxbn06dmm8r5o9gb79azrkvu6o.png)
But the problem is asking the answer in kilojoules, so remember that 1 kJ equals 1000 J, so the conversion will look like this:
![207328\text{ J}\cdot\frac{1\text{ kJ}}{1000\text{ J}}=207.33\text{ kJ.}](https://img.qammunity.org/2023/formulas/chemistry/college/upeg4lzrvwv8973mr99r6g0ebuc3wv6hv0.png)
The answer would be 207.33 kJ.