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

Answer:

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

Explanation:

$\Delta t=\frac{q}{m\cdot C_s}=\frac{5.83\times10^5}{2190\times 4.184}\\\\=63.6^\circ\:C$

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

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