chemistry

A piece of chocolate cake is burned with oxygen in a bomb calorimeter. The temperature of 3368 grams of H2O in the calorimeter is raised by 0.52 K. What is ∆E, including the sign, for the combustion of this piece ot chocolate cake? Assume no heat is absorbed by the calorimeter. The heat capacity of water is 1.0 cal/g · K and the heat of vaporization of water is 540 cal/g. Answer in units of kcal. doctorbob222

Answers



∆E = -55.6 kcal.

The change in energy of the system is equal to the heat absorbed by the water in the calorimeter. The heat absorbed by the water due to the combustion of the chocolate cake can be determined using the equation q = mass x specific heat capacity x change in temperature. This gives q = 3368 g x 1.0 cal/g·K x 0.52 K = 1749 cal. The heat of vaporization of water must also be considered, as the heat absorbed can cause the water to vaporize. The heat due to vaporization is given by q' = mass x heat of vaporization. This gives q' = 3368 g x 540 cal/g = 180252 cal. The total change in energy, ∆E, when the cake is burned is equal to the total heat absorbed, which is given by q + q' = 1749 cal + 180252 cal = 182002 cal, or 55.6 kcal (1 kcal = 4184 cal). Since the energy is going into the system, the answer is -55.6 kcal.

Answered by heatherpatterson

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