How much heat is needed to boil water on a camp stove?

[toothpaste666]

Homework Statement

The Pacific Crest Trail stretches from Mexico to Canada through the mountains of California, Oregon and Washington. Hikers on this trail must carefully estimate the amount of fuel they will need. They will typically carry a small camp stove which produces 700 J/s of heat (this is heat added to the food, extra heat is needed to heat the pan, stove, air, and ground, but we will ignore that) and which burns 0.15ℓ/h of fuel.

A)If the initial water temperature is 10◦C, how much time is required to bring 5ℓ of water to boiling temperature?

B)How much time will be required to melt 5 kg of ice which has an initial temperature of -5◦C? The final temperature will be 0◦C.

C)Water found in mountain streams is often boiled for 10 min to make sure that it is safe to drink. If the camp stove is used to boil 5ℓ of water for 10 min, how much water is vaporized?

Homework Equations

dQ = mcdT

The Attempt at a Solution

A) first i want 5 liters of water in kg
5 liters = 5E-3 m^3
d = m/v d for water = 1000kg/m^3
1000 = m /5E-3
m = .005(1000) = 5 kg
now use Q = mcdeltaT to find heat needed
where c the specific heat of water = 4186 J/kgC and the boiling temp of water = 100C
Q = mcdeltaT = (5kg) (4186 J/kgC) (100 C- 10 C) = 1883700 J
the stove burns 700J/s so
1883700J (s/700 J) = 2691 s to bring the water to boiling temp

B) heat of fusion of water/ice = 3.33E5 J/kg = L
specific heat of ice 2100 J/kgC
we need the heat to bring the water to 0C + heat to melt water
mcdeltaT + mL
5kg(2100 J/kgC) (0 - (-5 C)) + 5kg (3.33E5 J/kg)
= 5(2100)(5)J + 5(3.33E5)J = 1717500 J
which takes
1717500 J (s/700 J) = 2454 s

C) 10 min = 600 s. stove burns 700 J/s so
(700 J/s)(600 s) = 420000 J of heat are used
heat of vaporization for water L = 22.6E5 J/kg
the problem doesn't give a starting temp so i will assume 10 C
the stove will bring the water to 100C and then will vaporize some of it
let m be the total mass of water and let x be mass vaporized
Q = mcdeltaT + xL
420000 J = (5kg)(4186 J/kgC)(100 C -10 C) + x(22.6E5 J/kg)
420000 J = 1883700 J + x(22.6E5 J/kg)
-14636700 J = x (22.6E5 J/kg)
x = -.65 kg
it came out negative so no water would end up being vaporized? this is the part I am not so sure I did correctly

 

[haruspex]

the problem doesn't give a starting temp so i will assume 10 C

It doesn't say 'bring to the boil and boil for 10 minutes', so I think you should take it to be already at the boil at the start of the 10 mins.

 

[toothpaste666]

I thought that at first but then i thought i was wrong because wouldn't the set up be
Q = mL
but i already have Q and they give us 5liters which means m = 5kg and we know L also

 

[haruspex]

I thought that at first but then i thought i was wrong because wouldn't the set up be
Q = mL

Depends what you mean by m. You wrote, correctly:

Q = mc deltaT + xL

All I'm saying is deltaT =0.

 

[toothpaste666]

oh wow. they slipped that 5 liters in there as an almost red herring.
Q = xL
420000 J= x (22.6E5 J/kg)
x = .19 kg
thank you