Question

Calculate specific heat of liquid if a 450. g iron block is heated to 98.0 C and then droped into a calorimeter with 125 g of unknown liqid at 19.8 C. After 2 minutes the temp inside the calorimeter rise to a maximum of 40.4 C.

Answer 1

Specific heat of liquid
`1.258 \ J/g^0C.`

Step-by-step explanation:

We know in thermal equilibrium :

Loss in heat by iron block = Gain in heat by liquid .

Specific heat of iron = 0.45
`J/g^0C.` { source internet }

Now , loss in heat by iron block =
`mC_(iron)\Delta T=450* 0.45* 57.6=11664\ J.`

Heat gain by liquid=
`mC_(liquid)\Delta T=450* C_(liquid)* 20.6=9270* C_(liquid).`

Equating both we get :

`C_(liquid)=1.258 \ J/g^0C.`