Question #d1d51

Question

Question #d1d51

1 Answer

Impact of this question

2133 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-15T01:48:53

The initial temperature of the water is $39.9^{\circ} C$ $39.9 ""^@C$ .

Explanation:

$\frac{\frac{J}{1}}{\frac{J}{g} C} = g C$ $( J/1) / (J/gC) = gC$

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

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

$5.51 \times \frac{10^{5}}{4.184} / 2190 = 60.13^{\circ} C$ $5.51 xx 10^5 / 4.184 / 2190 = 60.13 ""^@C$ .

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

$100 . - 60.13 = 39.87^{\circ} C$ $100. - 60.13 = 39.87 ""^@C$

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

$39.87 = 39.9^{\circ} C$ $39.87 = 39.9 ""^@C$ to three significant digits.

Science

Math

Humanities

... and beyond

Question #d1d51

1 Answer

Explanation:

Related questions

Impact of this question