# Question d1d51

Sep 15, 2017

The initial temperature of the water is 39.9 ""^@C.

#### Explanation:

$\frac{\frac{J}{1}}{\frac{J}{g} C} = g C$

$g \times \frac{C}{g} = C$

Ao dividing the number of joules by the specific heat and then by the number of grams will give the value of the degrees ""^@C the water temperature has been raised.

 5.51 xx 10^5 / 4.184 / 2190 = 60.13 ""^@C.

This is the value of the number of degrees that the water has been heated so to find the initial temperature subtract the increase in temperature from 100 degrees the boiling point of water

 100. - 60.13 = 39.87 ""^@C

However since the number of joules $5.51 \times {10}^{5}$ has only three significant digits that have been measured the answer must be written in only three significant digits so

 39.87 = 39.9 ""^@C# to three significant digits.