Question

In: Physics

A 0.280-kg piece of aluminum that has a temperature of -166

Expert Solution

Amount of heat gained by aluminum = Amount of heat lost by water

Amount of heat gained by aluminum =

where c= specific heat of Aluminum = 900 J/kg C

m = mass of Aluminum = 0.280 kg

=change in temperature = 0 -(- 166) = 166 ( Since the equilibrium temperature os )

Therefore amount of heat lost by water is also 41832 J.

The water looses its heat in two steps. (i) In first step all the water cools down to (ii) In second step some part of water at converts in ice

(i) Amount of heat lost by water to reach

where c= specific heat of water = 4186 J/kg C

m = mass of water = 1.5 kg

=change in temperature = 3.1 - 0 = 3.1 ( Since the equilibrium temperature os )

By step (i) water has lost 19464.9 J . The remaining amount of heat ( 41832 J - 19464.9 J = 22367.1 J) will be lost to convert water into ice.

The laten heat of water is ( The laten heat of water is defined as the amount of heat lost by water to convert 1 gram of water to ice)

where m is mass of water converted to ice and L is latent heat

Therefore mass of water that has been frozen into ice is 66.97 gm ~ 67 gm

genius_generous answered 8 months ago

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