Find the temperature difference between the two bodies at time t.
Abhishek Dogra , 5 Years ago
Grade 11
AIIMS MBBS> Find the temperature difference between t...
1 Answers
Arun
Last Activity: 5 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)
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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