Question

Water (2190 g ) is heated until it just begins to boil. If the water absorbs 5.83×105 J of heat in the process, what was the initial temperature of the water?

Answer 1

The initial temperature was
`36.4^\circ \:C`

Step-by-step explanation:

`\Delta t=(q)/(m\cdot C_s)=(5.83*10^5)/(2190* 4.184)\\\\=63.6^\circ\:C`

The temperature difference
`=100-63.6=36.4^\circ\:C`

Best Regards!