Question

A 4.7-kg block of ice originally at 263 K is placed in thermal contact with a 16.2-kg block of silver (cAg = 233 J/kg-K) which is initially at 1067 K. The H2O - silver system is insulated so no heat flows into or out of it. At what temperature will the system achieve equilibrium?

Answer 1

The temperature will be "338 K".

Step-by-step explanation:

The given values are:

`c_(Ag)=233 \ J/kg-K`

Temperature,
`\Delta \ T=263 \ K`

As we know,

⇒
`Heat \ gain \ by \ ice = Heat \ last \ by \ silver`

`m_(ice)c_(ice) \Delta \ T+m_(ice)L_(ice)+m_(ice)c_(w) \Delta \ T=m_(Ag)c_(Ag) \Delta \ T`

On putting the values, we get

`Q_(ice)=4.7* 2030(273-263)+4.7(330* 10^3)+4.7* 4186(TK-273K)`

⇒
`=-3724646.6 \ J+19674.2 \ T`

`Q_(Ag)=16.7* 233(1087-T)`

⇒
`=4229625.7-3891.1 \ T`

Now,

System's final temperature will be:

`Q_(ice)=Q_(Ag)`

`-3724646.6 \ J+19674.2 \ T=4229625.7-3891.1 \ T`

`T=338 \ K`