Question

Water (2270 g ) is heated until it just begins to boil. If the water absorbs 5.65×105 J of heat in the process, what was the initial temperature of the water?Express your answer with the appropriate units.

Answer 1

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

`\begin{gathered} Q=mc\Delta T \\ \Delta T=(Q)/(mc) \end{gathered}`

Substitute the values in the above expression as

`\begin{gathered} \Delta T=\frac{5.65*10^5^{}}{(2270)(4.18)} \\ =59.54^0\text{C} \end{gathered}`

The intial temperature of the water is calculated as

`\begin{gathered} \Delta T=T-t \\ 59.54=100-t \\ t=100-59.54 \\ =40.46^0\text{C} \end{gathered}`

Hence, the initial temperature of the water is 40.46 degree Celsius