Thank you for the good advice.
Two
campers are desperately short of calor gas, yet they badly need a
hot drink.They estimate
that they have the equivalent of 1.12 dm3 of calor gas
in their “gas bottle “. Assume that the calor gas is pure butane
with a standard enthalpy change of combustion of –3000 kJ
mol-1.
Given
that the process is 77 % efficient, calculate the maximum mass of
water at a initial temperature of 20oC
which the campers could boil in order to make some hot coffee at
room temperature and pressure? Given that the specific heat
capacity of water is 4.2 J g-1 K-1
.
Solution: The
mass will be calculated uaing the relation q = mcT
step
I : For 'q' we need to find moles of butane as enthalpy of
combustion is given (3000)
n = 1.12 / 24 = 0.0466
enthalpy
of combustion must be converted to J as q gives value in
J
q =
3000,000 x 0.0466 = 139800 J
step
II: But the process is only 77% efficient hence q is:
139800 x 0.77 = 107646
step
III : finally we use q = mcT
107646 = m x
4.2x (T2-20)
What
to do next.......T2 is not given in the given or I am missing
something?