A 1.50 kg solid metal sphere at 100 oC is initially placed in a well-insulated v
ID: 513183 • Letter: A
Question
A 1.50 kg solid metal sphere at 100 oC is initially placed in a well-insulated vessel with 4.00 mol air at 20 oC, and the vessel is closed immediately. After some amount of time the sphere and the air both reach a temperature of 83.0 oC. Determine the heat capacity of the metal (in on/kg*oC) assuming that it is constant with temperature.
Ignore the green as that's a physics approach to answering the question. This is for chemical engineering where we use an energy balance and need to solve for the Cp of the metal. How would I go about doing that? What is written in pencil is the beginning of the approach that would be taken.
NAME: 3. (25) A 50 kg solid metal sphere at 1000c is placed in a ated vessel filled with 4.00 mol air at and the vessel is closed immediately. After some amount of time the sphere and the air reach a temperature of 83.00C. the heat capacity of the metal assuming that it is constant with temperature Note: Remember that this is a closed system and analyze accordingly. Initial Final 4.0 A mel Air A 11 Heat loot by metal Het gained laa ATair) A Tair 11) ummetne ATma e) ti, so yExplanation / Answer
Given
Air
No. of moles of air = 4 moles
Average mol. wt of air = 29 g/mol
mass of air = 4 mol * 29 g/mol = 116 g
Cp = 1.005 KJ/kg = 1.005 J/g.C (data for air)
Cp value doesnot vary much for 20 C and 83 C so Cp will remain constant
Intial temperature T1 = 20 C
Final temperature T2 = 83 C
Heat gained by air = m * Cp * ( T2 - T1) = 116 g * 1.005 J/g.C * ( 83 - 20) C = 7344.54 J
Metal
Mass = 1.5 Kg = 1500 g
Intial temperature T1 = 100 C
Final tempertaure T2 = 83 C
heat lost by metal = m * Cp * (T2 - T1) = 1500 g * Cp * ( 83 - 100 ) C = - Heat gained by air
1500 g * Cp * ( 83 - 100 ) C = - 7344..54 J
- 25500 g.C * Cp = - 7433.54 J
Cp = 0.288 J/g.C Answer
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