Question #d0725

1 Answer
Douglas K.
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 #

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