Question

An aluminum cylinder of a mass 푀QR=0.155kgis removed from a liquid nitrogen bath, where it has been cooled to 푇QR,T=−196°C. The cylinder is immediately placed in an insulated cup containing 푀X=0.0800kgof water at 푇X=15.0°C. The average specific heat of aluminum over this temperature range is 푐QR=653J/kg⋅K. The specific heat of liquid water is 푐X=4186J/kg⋅K. The latent heat of fusion for water is 퐿'=33.5×10JJ/kg. Is it possible for this system equilibrates at 0∘Cwithout any water ice forming?

Answer 1

No, it's no possible.

Step-by-step explanation:

For the system equilibrates, the heat lost by water must be equal to the heat gained by the aluminum. Let's suppose that the Ql for water is 0, it means that it will not change its physic phase, so:

Qw = Qa

mw *Cw*ΔTw = ma*Ca*ΔTa

Where m is the mass, c is the specific heat, and ΔT the temperature change (it can be used in ºC because of ΔºC = ΔK)

0.08*4186*(0ºC - 15ºC) = 0.155*653*(0ºC - (-196ºC))

-5,023.2 J = 19838.14 J

The values must be equal in the module only. So, it's impossible for the system equilibrates, the water will need to lose more heat, first by latent heat, becoming ice, and then it probably will need to cool down.