ok does this look right to anyone:
25gallons x 1.609 = 40.225km per gallon
150km/40.225 = 3.729 gallons to travel 150km
3.729 x 68% = 2.536kg
2.536 x 4.8x10^7J/kg = 1.22x10^9J ?
But don't I still need to know the type of metal or mass of metal to find the Q in order to find the deltaS for the metal?
And I don't even know how to find out the water part because what is 1.2L in kg?
what do you mean? little confused ... you said to use the m(i)*L for the left side of the equation and then for the right side, don't i use the specific heats for water and aluminum?
so i did what you said and this is what i got
(m(i)+2.0x10^-3kg)L = (m(w)*c(w)*deltaT(w) +(m(a)*c(a)*deltaT(a)
(m(i)+2.0x10^-3)(333) = (210x10^-3*4186*(12-0)) + (150x10^-3*900*(12-0))
m(i) = 36.54kg
this doesn't make sense does it??
Homework Statement
A piece of metal at 80C is placed into 1.2L of water at 72C. This thermally isolated system reaches a final temperature of 76C. Estimate the overall change of entropy for this system.
Homework Equations
deltaS = Q/T
Q=m*c*deltaT
The Attempt at a Solution
I...
Homework Statement
A block of ice at 0C is added to a 150g aluminum calorimeter cup which holds 210g of water at 12C. If all but 2.0 g of ice melt, what was the original mass of the block of ice
Homework Equations
heat gained by ice = heat lost by water + heat lost by calorimeter...
So i do 40.225 x 68% = 27.35 kg to get the mass of 25 gallons of gasoline?
and what next? I don't know a formula to figure out the energy given the information.
Homework Statement
Gasoline yields 4.8x10^7 J/kg when burned. How much energy does a car use traveling a distance of 150 km if it gets 25 miles per (British) gallon? Assume that the density of gasoline is 68% that of water.
Homework Equations
The Attempt at a Solution
I don't...