...temperature increase was observed? for 1.00 x 10^3 g of water. The entire candy bar weighed at 2.5 ounces. Calculate the fuel value (in nutrional Calories) of the sample and the total caloric content of the candy bar.
I used the formula of specific heat Q = m x T x SH. 1.00 x 10^3g h20 x 3.0 C x 1 cal/g h20 * C. g and C cancel out which leaves me with 1.00 x 10^3 x 3.0 X 1 cal = 3000 calories and then I divide by 1000 to get the nutrutional value which leaves me with 3 C or 3 kcal.
The answer is 2.1 x 10^2 kcal/candy bar. What am I doing wrong? Please help. Thanks
I used the formula of specific heat Q = m x T x SH. 1.00 x 10^3g h20 x 3.0 C x 1 cal/g h20 * C. g and C cancel out which leaves me with 1.00 x 10^3 x 3.0 X 1 cal = 3000 calories and then I divide by 1000 to get the nutrutional value which leaves me with 3 C or 3 kcal.
The answer is 2.1 x 10^2 kcal/candy bar. What am I doing wrong? Please help. Thanks