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 degres celsius is added to 0.20 kilograms of water at 60 degrees celsius. 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 degrees C) 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 degrees, 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