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A calorimeter, heat capacity H = 42.7 J K , contains 0.80 kg of water at 15C. ...
A calorimeter, heat capacity H = 42.7 J K , contains 0.80 kg of water at 15C. 0.40 kg of molten lead, specific heat capacity c = 157 J kg-1K-1and latent heat of fusion L = 2.31 x 10^4Jkg-1,is poured into the calorimeter. The final temperature of the system is 25.0 C. What was the initial temperature of the lead?
The specific heat capacity of water is 4182 J kg-1K-1
The melting temperature of lead is 327 C.
The specific heat capacity of solid lead is 136 J kg-1K-1.
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- First we have a sample of water (0.8kg) at 15C in a calorimeter.
- Then molten lead at T temperature is added in to the calorimeter.
- T>melting point of lead. So initially lead drops it temperature while it comes to melting point.
- When lead comes to melting point it uses its latent heat of fusion and become lead solid.
- When it become solid it will further decrease its temperature and the system will come to equilibrium
- At all these time water and calorimeter will absorb all the energy released by lead.
This question is bit complicated. We need to understand this step by step.
Let us first consider energy released by Lead
Heat released to become lead at melting point = Q1
Heat released at latent heat of fusion = Q2
Heat released up to equilibrium = Q3
`Q = mCtheta`
`Q1 = 0.4xx157xx(327-T)`
`Q = mL`
`Q2 = -0.4xx2.31xx10^4 = -4000J = -9.24KJ`
`Q = mcCtheta`
`Q3 = -0.4xx136xx(25-327) = -16428.8 = -16.43KJ`
(-) sign is used because the energy is released.
For absorbing energy
Energy absorbed by water = Q4
Energy absorbed by calorimeter = Q5
`Q = mCtheta`
`Q4 = 0.8xx4182xx(25-15) = 33456J = 33.46KJ`
`Q5 = Ctheta`
`Q5 = 42.7xx10 = 427J = 0.427KJ`
Energy released = energy absorbed
`-0.4xx157xx(327-T)/1000+9.24+16.43 = 33.46+0.427`
`T = 457.84`
So the initial temperature of lead is 454.84C
There is no heat losses in the system other than the releasing and absorbing heat between water,lead and calorimeter.
Posted by jeew-m on June 14, 2013 at 3:01 PM (Answer #1)
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