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

The value of specific heat of metal = 0.3537
`(KJ)/(Kg K)`

Step-by-step 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
`(KJ)/(Kg K)`

This is the value of specific heat of metal.