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A calorimeter, heat capacity H = 42.7  J K  , contains  0.80 kg of water at 15C. ...

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lak-86 | Student, Undergraduate

Posted June 14, 2013 at 1:56 PM via web

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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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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted June 14, 2013 at 3:01 PM (Answer #1)

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This question is bit complicated. We need to understand this step by step.

  • 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.

 

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

 

Assumption

There is no heat losses in the system other than the releasing and absorbing heat between water,lead and calorimeter.

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