Calculate Heat Value: Chemistry Problem Solved

[chemical]

ok...a burner containing 16.95g liquid fuel is used to heat 200g water in a beaker from 25 degrees to 44 degrees. assuming that half the heat produced is lost to its surroundings, calculate the heat value of the fuel in kg to the negative gram...

some chemistry problem but i can't find anywhere to do it

 

[ShawnD]

Originally posted by chemical
ok...a burner containing 16.95g liquid fuel is used to heat 200g water in a beaker from 25 degrees to 44 degrees. assuming that half the heat produced is lost to its surroundings, calculate the heat value of the fuel in kg to the negative gram...

some chemistry problem but i can't find anywhere to do it

LOL that is so damn easy simply because it is worded badly.
Look at what it is asking for "calculate the heat value of the fuel in kg to the negative gram".
It is telling me that my units are to be kg/g. Since there is always 1kg/1000g, my answer is simply 1kg/1000g


However, the question might actually want KJ/g so let's try that.
First, calculate the energy that the water gained.
Second, double that energy because only half of the total energy was put into the water.
Third, divide the total energy by the mass of the fuel in grams.

Scroll way down to compare your work with mine.



The total heat put into the water
E = mcT
E = (0.2Kg)(4.19KJ/Kg)(44-25)
E = 15.922kJ in the water

It says that only half of the heat produced was used in the water. That means
E = (2)(15.922)
E = 31.844KJ total

To get KJ/g, just divide those out.
31.844KJ/16.95g = 1.8787KJ/g
= 1.88KJ/g

 

[M98Ranger]

To solve this problem, we can use the formula Q = m x c x ΔT, where Q is the heat energy, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

First, we need to calculate the heat energy absorbed by the water. We know that the water's mass is 200g and the change in temperature is 44 degrees - 25 degrees = 19 degrees. Let's assume that the specific heat capacity of water is 4.18 J/g°C.

Q = (200g) x (4.18 J/g°C) x (19°C)
Q = 15,884 J

Since we know that half of the heat produced is lost to the surroundings, we need to divide the calculated heat energy by 0.5 to get the total heat energy produced by the burner.

Total heat energy produced = (15,884 J) / (0.5)
Total heat energy produced = 31,768 J

Next, we need to convert the mass of the fuel from grams to kilograms. 16.95g = 0.01695 kg

Now, we can use the formula for heat value, which is heat energy produced per unit mass of fuel.

Heat value = (Total heat energy produced) / (Mass of fuel)
Heat value = (31,768 J) / (0.01695 kg)
Heat value = 1,873,950 J/kg

Finally, we need to convert the heat value to the desired units of kg to the negative gram. To do this, we need to divide the value by 1000 (to convert from J to kJ) and then multiply by -1 (to get the negative exponent).

Heat value = (1,873,950 J/kg) / (1000 J/kJ) x (-1)
Heat value = -1.87395 kg^-1 g

Therefore, the heat value of the fuel is -1.87395 kg^-1 g.