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
To calculate ∆E for the combustion of the chocolate cake, we need to use the equation:
∆E = q + w
Where:
∆E is the change in internal energy
q is the heat transferred to or from the system
w is the work done on or by the system
In this case, since the cake is burned in a bomb calorimeter, no work is being done (w = 0), so the equation simplifies to:
∆E = q
We can calculate the heat transferred (q) using the equation:
q = m × C × ∆T
Where:
m is the mass of the water
C is the specific heat capacity of water
∆T is the change in temperature
Given:
Mass of water (m) = 3368 grams
Specific heat capacity of water (C) = 1.0 cal/g · K
Change in temperature (∆T) = 0.52 K
Plugging in these values into the equation, we have:
q = (3368 g) × (1.0 cal/g · K) × (0.52 K)
q = 1755.04 cal
Now, since the question asks for the answer in kcal, we divide the result by 1000 to convert from calories to kilocalories:
∆E = q / 1000
∆E = 1755.04 cal / 1000
∆E = 1.75504 kcal
Therefore, the ∆E for the combustion of the chocolate cake is approximately 1.75504 kcal.