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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 .]
[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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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
Most Upvoted Answer
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
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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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? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about 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? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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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