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In the temperature range 100-1000 C, the molar specific heat of a metal varies with temperature T (measured in degrees Celsius) according to the formula Cp =(1 + T /5 ) J-deg C-1-mol-1. If 0.2 Kg of the metal at 300 C, the final equilibrrium temperature, in degree C, will be ________ .
[Assume that no heat is lost to radiation and /or other effects .]
Correct answer is '519'. Can you explain this answer?
Verified Answer
In the temperature range 100-1000 C, the molar specific heat of a meta...

m1 = 0.2. T1= 600 C 
m2 = 0.1, T2 = 300 C
We know at the thermal equilibrium 
Heat absorb = Heat gain


3T2 + 30T = 15000 + [3002 + 2(600)2]  
T = 519
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In the temperature range 100-1000 C, the molar specific heat of a meta...
Solution:

Given, Cp = (1 - T/5) J-deg C-1-mol-1
Let the metal be in equilibrium at temperature T1
The heat gained by the metal, Q1 = mCp(T1 - 300)
At equilibrium temperature T2, the heat lost by the metal, Q2 = mCp(T2 - T1)
Since the metal is in equilibrium, Q1 = Q2
mCp(T1 - 300) = mCp(T2 - T1)
Simplifying, we get T2 = 2T1 - 300
To find T1, we can use the fact that the metal is in equilibrium, i.e., Q1 = Q2
mCp(T1 - 300) = mCp(T1 - T2)
Substituting T2 = 2T1 - 300, we get
mCp(T1 - 300) = mCp(T1 - (2T1 - 300))
Simplifying, we get T1 = 519 C

Answer: 519 C
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In the temperature range 100-1000 C, the molar specific heat of a metal varies with temperature T (measured in degrees Celsius) according to the formula Cp =(1 + T /5 ) J-deg C-1-mol-1. If 0.2 Kg of the metal at 300 C, the final equilibrrium temperature, in degree C, will be ________ .[Assume that no heat is lost to radiation and /or other effects .]Correct answer is '519'. Can you explain this answer?
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