Question

A 140.0-g sample of water at 25.0°C is mixed with 108.6 g of a certain metal at 100.0°C. After thermal equilibrium is established, the (final) temperature of the mixture is 29.6°C. What is the specific heat capacity of the metal, assuming it is constant over the temperature range concerned?

Answer 1

Answer:

The value of specific heat of metal = 0.3537 $\frac{KJ}{Kg K}$

Explanation:

Mass of water $m_{w}$ = 140 gm = 0.14 kg

Initial temperature of water $T_{w}$ = 25°c = 298 K

Mass of metal $m_{metal}$ = 108.6 gm = 0.1086 kg

Initial temperature of metal $T_{metal}$ = 100°c = 373 K

After thermal equilibrium the final temperature  $T_{f}$ = 29.6°c = 302.6 K

Apply energy principal

Heat lost by metal = heat gain by water  --------- (1)

⇒ Heat lost by metal = $m_{metal}$ × $C_{metal}$ × ( $T_{metal}$ - $T_{f}$ )

⇒ Heat lost by metal = 0.1086 × $C_{metal}$ × ( 373 -302.6 )

⇒ Heat lost by metal = $C_{metal}$ × 7.64544  KJ -------- (2)

Now

Heat gain by water = $m_{w}$ × $C_{water}$ × ( $T_{f}$ - $T_{w}$ )

⇒ Heat gain by water = 0.14 × 4.2 × ( 302.6 - 298)

⇒ Heat gain by water = 2.7048 KJ --------------------- (3)

From the energy principal

Equation 2 = equation 3

$C_{metal}$ × 7.64544  = 2.7048

$C_{metal}$ = 0.3537 $\frac{KJ}{Kg K}$

This is the value of specific heat of metal.