A calorimeter containes 50g of water at `50^(@)C` . The temperature falls to `45^(@)C` in 10 minutes. When the calorimeter contains 100g of water at `50^(@)C` it takes 18 minutes for the temperature to become `45^(@)C` . Find the water equivalent of the calorimeter.

Answer

A, B

Explanation

`50^@C … 45^@C … 4^@C`
Let the surrounding temperature be `T^@C`
Average `T = (50 + 45)/(2)`
`95/2 = 47.5`
Average temperature difference from surrounding
`T = 47.5 - t`
`(50 - 45)/(5) = 1^@C//m`.
Rate of fall of temperature `= 5`
From Newton's law
`1^@C//mm = bA xx T`
`implies bA = 1/T = 1/(47.5 - t)`
In 2nd case,
Average temperature
`= (40 + 45)/(2) = 85/2 = 42.5`
Average temperature difference from surrounding
`T' = 42.5 - t`
Rate of fall of temperature
`= (45 - 40)/(8) = (5/8) ^@C//m`
From Newton's law
`5/8 = bAT'`
`implies 5/8 = (42.5 - t)/(47.5 - t)`
Hence, `t = 34.1 ^@C`.