Physics, asked by saahithi41, 1 year ago

cooling rate of sphere of 600K at external environment (200K) is R. when the temperature of sphere is reduced to 400K then cooling rate of the sphere becomes

Answers

Answered by chandresh126
64

Answer :

Cooling rate of the sphere becomes \frac{3}{16} H

Step by Step Explanation :


Rate of cooling ∝ (T⁴-T₀⁴) Þ \frac{H}{H'}


=> (\frac{T^{4}_{1} -T^{4}_{0}}{T^{4}_{2}-T^{4}_{0}}=\frac{400^{4}-200^{4}}{600^{4}-200^{4}}


=> H′ = \frac{ (16+4)(16-4)H}{(36+4)(36-4)}


=> \frac{3}{16} H

Answered by KishoreRishwanth
0

Explanation:

Cooling rate of the sphere becomes \frac{3}{16} H

16

3

H

Step by Step Explanation :

Rate of cooling ∝ (T⁴-T₀⁴) Þ \frac{H}{H'}

H

H

=> (\frac{T^{4}_{1} -T^{4}_{0}}{T^{4}_{2}-T^{4}_{0}}=\frac{400^{4}-200^{4}}{600^{4}-200^{4}}

T

2

4

−T

0

4

T

1

4

−T

0

4

=

600

4

−200

4

400

4

−200

4

=> H′ = \frac{ (16+4)(16-4)H}{(36+4)(36-4)}

(36+4)(36−4)

(16+4)(16−4)H

=> \frac{3}{16} H

16

3

H

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