Final temperature of water

Question

The amount of energy released by burning a fuel source, measured in energy per mass, is called the fuel value. If all of the energy obtained from burning 1.27 lb of ethane (fuel value is 11.36) is used to heat 128.0 kg of water at an initial temperature of 19.4°C. What will the final temperature of the water be? 

Answer 1

You neglected to provide the units of the fuel value for ethane. The value of 11.36 should have units of kcal/g. With that additional information, we can now attempt to answer the question.

First, we will convert 1.27 lb ethane to grams ethane:

1.27 lb x 454 g / lb = 576.6 g ethane

Heat generated by 575.6 g ethane = q:

q = 576.6 g ethane x 1 kg / 1000 g x 11.36 kcal/kg = 6.550 kcal of heat

Now, we can use q = mC∆T to find the final temperature of the water:

q = heat = 6.550 kcal

m = mass = 128.0 kg water

C = specific heat of water = 1 kcal/kgº

∆T = change in temperature = ?

Solving for ∆T, we have..

∆T = q/mC = 6.550 kcal / (128 kg)(1 kcal/kgº)

∆T = 0.051º

Final temperature = 19.4º + 0.051º = 19.5ºC

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Using the more conventional 4.184 J/gº as specific heat of water, and units of joules (J) for heat, we can re-do the calculations:

q = heat = 6.55 kcal x 1000 cal / kcal x 4.184 J/cal = 27405 J of heat generated from the ethane

For the water:

q = mC∆T

27405 J = (128,000 g)(4.184 J/gº)(∆T)

∆T = 0.051º

Final temperature = 19.4º + 0.05º = 19.5º