Question

Water (2850 g) is heated until it just begins to boil. If the water absorbs 5.45*10^5 of heat in the process, what was the initial temperature of the water?

Answer 1

Heat Capacity of water is 4.186 J/g C

Heat absorbed = Heat capacity * MAss of water * (Final Temp. - Initial Temp)

As the water starts to boil Final Temp = 100 C

Given, Heat absorbed = 5.45*10⁵ J

Put all these in equation above

5.45*10⁵ J = 4.186 J/g C * 2850g * (100-Initial Temp)

which reduces to

45.68C = (100-Initial Temp)

So, Initial Temp = 54.32C (Answer)