Melting ice
1. What would be the
temperature for the boiling point of water on a new scale designed so that the melting point is
25 degrees and human body temperature is 150 degrees?
2. A 0.35 kilogram of ice
at -15 oC is added to 0.20 kilograms of water at 60 oC. How much
water is there when the ice and water reach thermal
equilibrium?
Answer:
1. I assume that the melting point that you
mention is that of ice (0 oC) and that the human body temperature is 37
oC.
So a difference in C of 37 degrees is equivalent to a different of 125
degrees on the new scale.
Therefore 1 degree C is equivalent to 125/37 = 3.38 degrees
on the new scale.
A 100 degrees C is equivalent to 338 degrees on the new scale
So
the boiling point of water (100 degrees C) = 25 + 338 = 363 o C
2. For this
one I will need to assume some data :
Specific latent heat of fusion of ice = 334 000
J/kg
Specific heat capacity of water = 4200 J/kgK
Specific heat capacity of ice =
2100 J/kgK
Heat energy lost by ice = heat energy gained by water
We
assume that the final temperature of the ice water mixture is 0 oC, in other words some
of the ice will have melted but not all of it.
0.35x2100x15 + 334 000 x m =
0.2x4200x60
where m is the mass of ice that has melted
11025 +
334000m = 50400
50400 – 11025 = 334000m giving m = 0.12 kg
So the mass
of water at equilibrium = 0.2 + 0.12 = 0.32 kg