A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 59.7 ∘C. Part A What is the mass of the water? Express your answer to two significant figures and include the appropriate units.

The mass of water = 39.18 gm

Explanation:

Mass of iron = 32.5 gm

Initial temperature of iron = 22.4°c = 295.4 K

Specific heat of iron   = 0.448

Mass of water =

Specific heat of water  

Initial temperature of water = 336 K  

Final temperature after equilibrium = 59.7°c = 332.7 K

When iron rod is submerged into water then

Heat lost by water  = Heat gain by iron rod

( - ) =   ( - )

Put all the values in above formula we get

× 4.2 × ( 336 - 332.7 ) = 32.5 × 0.448 × ( 332.7 - 295.4 )

= 39.18 gm

Therefore the mass of water = 39.18 gm