Question #d0725

1 Answer
Dec 28, 2016

The specific heat of water is (4.1868" Joules")/("g "^@"C). The change in temperature is 80^@"C". Please see the explanation for the factor-label calculation.

Explanation:

(210 cancel("g of H"_2"O"))/1(4.1868" Joules")/(cancel("g "^@"C of H"_2"O"))(80cancel(@C))/1(1/(105" s")) ~~ 670" Joules"/s