A cylindral copper rod of diameter 2m and length 20cm is heated at one end by steam at 100°c . The other end is kept at 0°c . Find the quantity of heat conducted across any section per minute if thermal conductivity of copper is 385Wm-1 K-1.
Answers
Answer:
Explanation:
Basic equation connecting thermal conductivity extrinsic factors like heat flow, temperature gradient area and length is
Rate of heat flow = Thermal Conductivity x Temp Difference × Area /Length
Q = k × Δ T × A L ......(1)
Note that .
Q is in Watts or Calory per second; it is power, or energy per unit time.
Heat Q required to melt 10 grams of ice = m × L
where L is latent heat of fusion of water and is = 79.7 c a l ⋅ g −1
Q = 10 × 79.7 = 797 c a l
Thermal conductivity of Copper
k = 0.99 c a l ⋅ s − 1 c m − 1⋅
K − 1
Inserting given values in equation (1) we get .
Q = 0.99 × ( 100 − 0 ) ×π × 1 2 120 ⇒ . Q = 2.59 c a l ⋅ s − 1
Number of seconds required to conduct
797 c a l of heat = 797 2.59 = 308 s
, rounded to last second
Answer:
the answer to your specific question is
Explanation:
Number of seconds required to conduct
797 c a l of heat = 797 2.59 = 308 s
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