Physics Thermal Properties of Matter
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New answer posted
a month agoContributor-Level 10
R_eq = R? + R?
L_total / (K_eq * A) = L? / (K? A) + L? / (K? A)
Assuming L? = L? = l, L_total = 2l
2l / (K_eq * A) = l/ (K? A) + l/ (K? A)
2/K_eq = 1/K? + 1/K? = (K? + K? )/ (K? )
K_eq = 2K? K? / (K? + K? )
New answer posted
a month agoContributor-Level 10
In steady state rate of flow of heat in all three rods are same.
dQ/dt = (k? A (100 – 70)/? = (k? A (70 – 20)/? = (k? A (20 – 0)/?
30k? = 50k? = 20k?
∴ k? : k? = 2:3 & k? : k? = 2:5
New question posted
a month agoNew answer posted
a month agoContributor-Level 10
At T°C L = L? + L?
At T +? T Leq = L'? + L'?
where L'? = L? (1 + α? T)
L'? = L? (1 + α? T)
Leq = (L? + L? ) (1 + αavg? T)
⇒ (L? + L? ) (1 + αavg? T) = L? + L? + L? α? T + L? α? T
⇒ (L? + L? )αavg = L? α? + L? α?
⇒ αavg = (L? α? + L? α? )/ (L? + L? )
New answer posted
a month agoContributor-Level 10
The decrement in length is more for metal strip-A than metal strip-B, so the combined system bend towards the left.
New answer posted
a month agoContributor-Level 9
dQA = mACA dTA
dQB = mBCB dTB
dQA/dt = dQB/dt
⇒ mACA (dTA/dt) = mBCB (dTB/dt)
⇒ (dTA/dt)/ (dTB/dt) = (mBCB)/ (mACA)
Assuming mA = mB
(dTA/dt)/ (dTB/dt) = CB/CA
CA/CB = (dTB/dt)/ (dTA/dt) = (90/6)/ (120/3) = 15/40 = 3/8
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