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Find the temperature difference between the two bodies at time t.

Find the temperature difference between the two bodies at time t.

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Grade:11

1 Answers

Arun
25758 Points
3 years ago
Let T₁, T₂ be the temperatures of object 1 and object 2 respectively and T₁₀ and the T₂₀ corresponding initial values. 

Let Object 2 transfer heat at flow rate H=dQ/dt, thus is rate of change of temperature is  
given by 
m₂∙s₂∙(dT₂/dt) = -H 
 
dT₂/dt = - H/(m₂∙s₂) 
Since connecting rod has not heat capacity is cannot store no energy. Therefore at same instant the same amunt of heat is absorbed by object 1. Thus, 
m₁∙s₁∙(dT₁/dt) = + H 
 
dT₁/dt = H/(m₁∙s₁) 

Because rod stores not heat we can assume quasi-stationary heat flow and apply Fourier's law in one-dimensional from. So heat flow between the objects is given by: 
H = (K∙A/L)∙(T₂ - T₁)  

Now introduce the temperature difference 
ΔT = T₂ - T₁ 

Hence; 
dΔT/dt = dT₂/dt - dT₁/dt = - H/(m₂∙s₂) - H/(m₁∙s₁) = - H∙(1/(m₂∙s₂) - 1/(m₁∙s₁)) 
 
dΔT/dt = - (K∙A/L)∙(T₂ - T₁)∙(1/(m₂∙s₂) - 1/(m₁∙s₁)) = - (K∙A/L)∙(1/(m₂∙s₂) - 1/(m₁∙s₁))∙ΔT 
 
dΔT/dt = - k∙ΔT 
where k = (K∙A/L)∙(1/(m₂∙s₂) - 1/(m₁∙s₁))∙ΔT 

To solve this differential equation separate variables and integrate 
(1/ΔT) dΔT = - k dt 
=> 
∫ (1/ΔT) dΔT = ∫ - k dt 
=> 
ln(ΔT) = -k∙t + c 
where c is the constant of integration 
 
ΔT = e^( -k∙t + c) = -k∙t + c = e^(c)∙e^(-k∙t) = C∙e^(-k∙t) 
where C = e^(c) 

at t=o the temperature difference is: 
ΔT₀ = T₂₀ - T₁₀  
that means 
ΔT₀ = C∙e^(-k∙0) = C 

So the temperature difference as function of time is given by 
ΔT = ΔT₀∙e^(-k∙t) 
= (T₂₀ - T₁₀)∙e^(- (K∙A/L)∙(1/(m₂∙s₂) - 1/(m₁∙s₁)) ∙ t)

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