A new linear temperature scale designed for specific experiment records the ice point at 22°T (°T is the new unit of temperature) and the steam point 232°T. If specific heat capacity of water is 4200 in `J kg^(-1) ""^(@)C^(-1)` ,then what will be its value in `J kg^(-1) ""^(@)T^(-1)` ?

Answer

2000

Explanation

(Steam point - ice point) = `100^(@)` C
`therefore (232 -22) ""^(@)T = 100^(@) C`
`therefore 210^(@) T = 100^(@)C`
`rArr 1^(@) C = 2.1^(@) T`
Specific heat of water = 4200 J `kg^(-1) ""^(@)C^(-1)`
`= 4200/2.1 J kg^(-1) ""^(@)T^(-1)`
`= 2000 J kg^(-1) ""^(@)T^(-1)`