Question

A piece of iron (mass = 25.0 g) at 398 K is placed in a styrofoam coffee cup containing 25.0 mL of water at 298 K. Assuming that no heat is lost to the cup or the surroundings, what will the final temperature of the water be? The specific heat capacity of iron = 0.449 J/g°C and water = 4.18 J/g°C.

Answer 1

Answer:

This is the final temperature of water = 398 K

Explanation:

Mass of iron $m_{iron}$ = 25 gm

Initial temperature of iron $T_{iron}$ = 398 K

Mass of coffee cup $m_{c}$ = 25 ml = 25 gm

Temperature of coffee cup $T_{c}$  = 298 K

Specific heat of iron $C_{iron}$ = 0.449 $\frac{J}{g c}$

Specific heat of water $C_{water}$ = 4.18 $\frac{J}{g c}$

From the energy balance principal,

Heat lost by the iron = heat gain by the coffee cup

⇒ $m_{iron}$ × $C_{iron}$  × $(T_{iron} - T_{f} )$ = $m_{c}$ × $C_{cof}$ × $(T_{f} - T_{c} )$

⇒ 25 × 0.449 × $(398 - T_{f} )$ = 25 × 4.18 ×  $( T_{f} - 398 )$

⇒ 0.449 × $(398 - T_{f} )$ =  4.18 ×  $( T_{f} - 398 )$

⇒ 398 - $T_{f}$ = 9.3  ( $T_{f}$ - 398 )

⇒ $398 - T_{f}$ = 9.3 $T_{f}$ - 3701.4

⇒ 10.3 $T_{f}$ = 4099.4

⇒ $T_{f}$ = 398 K

This is the final temperature of water.