How do I find the change in temperature of $199.0 g$ $"199.0 g"$ of water that loses $2793 J$ $"2793 J"$ of heat?

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Answer 1 · 2017-09-15T01:37:38

It goes down by 3.354 degrees C.

$\frac{J}{\frac{J}{g \cdot^{\circ} C}} = g^{\circ} C$ $"J"/("J"/("g"cdot""^@ "C")) = "g"^@ "C"$

$\frac{g^{\circ} C}{g} =^{\circ} C$ $("g"^@"C") / "g" = ""^@"C"$

So dividing the number of joules 2793 by the specific heat and then by the number of grams will give the change in degrees C:

$\frac{2793 J}{{4.184 J/g}^{\circ} C} = {667.5 g}^{\circ} C$ $"2793 J"/("4.184 J/g"^@ "C") = "667.5 g"^@"C"$

${667.5 g}^{\circ} C / 199.0 g = {3.354}^{\circ} C$ $"667.5 g"^@"C" // "199.0 g" = 3.354^@ "C"$ .

How do I find the change in temperature of 199.0 g"199.0 g" of water that loses 2793 J"2793 J" of heat?