A32.5 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?

the mass of the water is 29.4 g

Explanation:

given information:

mass of iron, = 32.5 g

iron's temperature, = 21.8°C

water's temperature, = 63.3°C

thermal equilibrium, = 58.9°C

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

= - , heat being transfer

Δ =tex]m_{w}[/tex] Δ

(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