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Rate of cooling of body is 0.5^(@)C/min,...

Rate of cooling of body is `0.5^(@)C`/min, when the system is `50^(@)C` above the surroundings. When a system is `30^(@)C` above the surroundings, the rate of cooling will be

A

`0.3^(@)C`/min

B

`0.6^(@)C`/min

C

`0.7^(@)C`/min

D

`0.4^(@)C`/min

Text Solution

Verified by Experts

The correct Answer is:
A
`(R_(2))/(R_(1))=(K(theta_(2)-theta_(1)))/(K(theta_(1)-theta_(0)))=(30)/(50)`
`R_(2)=(3)/(5)R_(1)`
`=(3)/(5)xx0.5=0.3.^(@)C//`min.
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