The temperature of 500.0 g of water contained in a calorimeter increased from 20.0 °c to 50.0 °c when 5.00 g of fuel was burned to completion. what is the δh of this combustion reaction? (the specific heat of water is 4.184 j/g°c and the molecular weight of the fuel is 40.0 g/mol.)

The ΔH of this combustion reaction is -502 kJ/mol

Explanation:

Step 1: Data given

Mass of water = 500.0 grams

Mass of fuel = 5.00 grams

Initial temperature = 20.0 °C

Final temperature = 50.0°C

The specific heat of water is 4.184 J/g°C

Molecular weight of the fuel is 40.0 g/mol

Step 2: Calculate the heat transfer

q = m*c* ΔT

⇒ with m = the mass of water = 500 grams

⇒ with c = the specific heat of water = 4.184 J/g°C

⇒ with ΔT = the change in temperature = T2 - T1 = 50.0°C- 20.0°C  = 30.0°C

q = 500 * 4.184 * 30

q= 62760 J

Step 3: Calculate moles

moles = mass / molar mass

Moles = 5 grams / 40 g/mol

moles = 0.125 moles

Step 4: Calculate  ΔH of this combustion reaction

ΔH = 62760 / 0.125 mol

ΔH = 502080 J/mol = 502 kJ /mol ( negative because it's an exothermic reaction)

ΔH = -502 kJ/mol

i would say b

explanation:

but i think i would need more information than that

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.