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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?

Respuesta :

Answer:

The initial temperature was  [tex]36.4^\circ \:C[/tex]

Explanation:

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

The temperature difference [tex]=100-63.6=36.4^\circ\:C[/tex]

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