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

Answer:

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

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 = $mC_{iron}\Delta T=450\times 0.45\times 57.6=11664\ J.$

Heat gain by liquid=$mC_{liquid}\Delta T=450\times C_{liquid}\times 20.6=9270\times C_{liquid}.$

Equating both we get :

$C_{liquid}=1.258 \ J/g^0C.$