Question

A 32.5 g iron rod, initially at 21.8°C, is submerged into an unknown mass of water at 63.3°C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 58.9°C. What is the mass of the water?

Answer 1

Answer: Mass of water is 29.50g

Step-by-step explanation:

Since the system is well insulated the heat lost to the surrounding is assumed to be zero(0)

Therefore,

Heat gained by iron = Heat lost by the water

Mi×Ci×∆Ti = Mw×Cw×∆Tw .... eqn1

Where Mi and Mw = mass of iron and water respectively

Ci and Cw = specific heat capacity of iron and water respectively

∆Ti and ∆Tw = change in temperature of iron and water respectively

Mi= 32.5g, Mw= ? , Ci = 0.45J/gC, Cw= 4.18J/gC,

∆Ti = 58.9-21.8 = 37.1°C

∆Tw = 63.3-58.9 = 4.4°C

From eqn1

Mw = (Mi×Ci×∆Ti)/(Cw×∆Tw)

Mw = ( 32.5 × 0.45 × 37.1)/ ( 4.18 × 4.4)

Mw = 29.50g

Answer 2

the mass of the water is 29.4 g

Step-by-step explanation:

given information:

mass of iron,
`m_(i)` = 32.5 g

iron's temperature,
`T_(i)` = 21.8°C

water's temperature,
`T_(w)` = 63.3°C

thermal equilibrium,
`T_(eq)` = 58.9°C

since water and iron is being mixed, q of the water and the iron are equal, thus

`q_(i)` = -
`q_(w)`, heat being transfer

`m_(i)`
`C_(i)` Δ
`T_(i)` =tex]m_{w}[/tex]
`C_(w)` Δ
`T_(w)`

(32.5)(0.449)(58.9-21.8) = tex]m_{w}[/tex] (4.18)(58.9-63.3)

tex]m_{w}[/tex] = (32.5)(0.449)(58.9-21.8)/(4.18)(58.9-63.3)

= 29.4 g