How much alcohol must be burned to generate 279kj of heat to warm the water as described in
problem 3?(from 60 F to 180 F)
a. 18.7 g
b. 2.9 g
c. 1.6 g
d. 10.4 g
e. 9.3 g
the answer is D help with solution?
You need the given heat of combustion for the alcohol.
massAlc*HeatCombustion= 279kj
solve for mass of Alcohol.
how would i know which heat of combustion to use for this question?
To solve this problem, we can use the principle of energy conservation, specifically the equation for heat transfer:
Q = mcΔT
where Q is the heat transferred, m is the mass of the substance being heated, c is the specific heat capacity of the substance, and ΔT is the change in temperature.
We have the following information:
- Q = 279 kJ (converted to J)
- Initial temperature (Ti) = 60°F (converted to °C)
- Final temperature (Tf) = 180°F (converted to °C)
First, let's calculate the mass of water that is being heated. We assume that water has a specific heat capacity of 4.18 J/g°C:
Q = mcΔT
To convert the mass from grams to kilograms, we divide the mass by 1000:
m = Q / (cΔT)
m = (279,000 J) / (4.18 J/g°C * (180°C - 60°C))
m = (279,000 J) / (4.18 J/g°C * 120°C)
m ≈ 581.56 g
Next, we need to calculate the amount of alcohol that must be burned to generate this amount of heat. The heat of combustion of alcohol depends on the specific alcohol being burned. Let's assume it is ethanol (C2H5OH).
The heat of combustion for ethanol is approximately 30 kJ/g.
To find the mass of alcohol, we can use the equation:
Q = mc
Where Q is the heat generated, m is the mass of alcohol, and c is the heat of combustion of alcohol.
m = Q / c
m = (279,000 J) / (30,000 J/g)
m = 9.3 g
Therefore, the answer is option e) 9.3 g.