How are the ?H values for reactions determined in the first place? The reactions

Question

How are the ?H values for reactions determined in the first place? The reactions are run in a calorimeter, which restricts and defines the surroundings so that thermal energy transfer to or from the surroundings (prompted by a change in temperature ?T) can be easily measured. Why measure the thermal energy change of the surroundings? So that we can then determine the thermal energy change of the system. Remember that thermal energy change, or heat transfer q , of the system and surroundings can be expressed as:

qsystem + qsurroundings = 0 (energy neither created nor destroyed)

qsystem = ? qsurroundings

One other thing: if your system is a tangible thing with a specific heat capacity, and your surroundings are tangible with a specific heat capacity, you can use the following:

qsystem = (m s ?T)system   and   qsurroundings = (m s ?T)surroundings    

Substitute this into qsystem = ? qsurroundings to get:     

(m s ?T)system = ?(m s ?T)surroundings

What mass of a hot (85.0

Explanation / Answer

we know that

mass = density x volume

given

density of water = 1 g/ml

volume of water = 200 ml

so

mass of water = 1 x 200

mass of water = 200 g

now

Heat = mass x specific heat x temp change

given

specific heat of water = 200 g

temp change of water = 30 -25 = 5 C

so

Heat = 200 x 4.184 x 5

Heat = 4184 J

so

heat required to increase the temp of water is 4184 J

now this heat comes from aluminium

so

Heat = mass x specific heat x temp change

the final of the system is 30 C

so the final temp of aluminium is 30 C

so

temp change = 30-85 = -55

specific heat of aluminium = 0.9

so

Heat released from aluminium = m x s x dT

-4184 = m x 0.9 x -55

m = 84.52 g

so

84.52 g of aluminium rod need to be placed

simple method :

heat lost by hot body = heat gained by cold body

here water is the cold one and aluminium is hot

so

mal x sal x dTal = mw x sw x dTw

mal x 0.9 x (85-30) = 200 x 4.184 x 5

mal = 84.5 g

so

84.52 g of aluminium rod need to be placed

so the answer is option c) 84.5 g

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