Question

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.

Answer 1

Answer:

The mass of water $m_{w}$ = 39.18 gm

Explanation:

Mass of iron $m_{iron}$ = 32.5 gm

Initial temperature of iron $T_{1}$ = 22.4°c = 295.4 K

Specific heat of iron  $C_{iron}$ = 0.448 $\frac{KJ}{kg K}$

Mass of water = $m_{w}$

Specific heat of water  $C_{w} = 4.2 \frac{KJ}{kg K}$

Initial temperature of water $T_{2}$ = 336 K  

Final temperature after equilibrium $T_{f}$ = 59.7°c = 332.7 K

When iron rod is submerged into water then

Heat lost by water  = Heat gain by iron rod

$m_{w}$ $C_{w}$ ($T_{2}$ - $T_{f}$ ) =  $m_{iron}$ $C_{iron}$ ( $T_{f}$ - $T_{1}$ )

Put all the values in above formula we get

$m_{w}$ × 4.2 × ( 336 - 332.7 ) = 32.5 × 0.448 × ( 332.7 - 295.4 )

$m_{w}$ = 39.18 gm

Therefore the mass of water $m_{w}$ = 39.18 gm