The rate of cooling of a body is `0.5^(@)C//"minute"`, when the body is `50^(@)C` above the surrounding. What is the rate of cooling if the body is `30^(@)C` above the surrounding ?

Options

A. `24^(@)C//"hour"`
B. `18^(@)C//"hour"`
C. `30^(@)C//"hour"`
D. `12^(@)C//"hour"`

Answer

B

Explanation

According to Newton's law of cooling
`(d theta)/(dt)=K[theta-theta_(0)]=K` (temperature difference)
`therefore0.5=K(50)" "thereforeK=0.5/50=10^(-2)`
If the temperature difference is `30^(@)C`, then `(d theta)/(dt)`
`=K(30)=10^(-2)xx30=0.3^(@)C//"minute"`
`therefore` Rate of cooling per hour = `0.3xx60=18^(@)C//"hour"`