Given data
*The given mass is m = 2270 g
*The amount of heat is Q = 5.65 × 10^5 J
*The final temperature of the water is T = 100 C
*The specific heat capacity of water is c = 4.18 J/g C
The formula for the amount of heat required is given as
[tex]\begin{gathered} Q=mc\Delta T \\ \Delta T=\frac{Q}{mc} \end{gathered}[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} \Delta T=\frac{5.65\times10^5^{}}{(2270)(4.18)} \\ =59.54^0\text{C} \end{gathered}[/tex]The intial temperature of the water is calculated as
[tex]\begin{gathered} \Delta T=T-t \\ 59.54=100-t \\ t=100-59.54 \\ =40.46^0\text{C} \end{gathered}[/tex]Hence, the initial temperature of the water is 40.46 degree Celsius
