3/20
101 over 102 is already in simplest form f
To convert water at 200°F to ice at 30°F, you need to remove 1 BTU to cool water from 200°F to 32°F to become ice. Then, you need to remove 144 BTUs to cool the ice from 32°F to 30°F. So, total BTUs needed to remove from one pound of water at 200°F to end up as ice at 30°F is 144 + 1 = 145 BTUs.
To convert 1 lb of water at 200°F to ice at 30°F, you need to remove 233 BTUs. This involves cooling the water from 200°F to 32°F (latent heat of fusion) and then further cooling the ice from 32°F to 30°F.
200
actually its 313.
50-200 USD
313 Btu
To change water at 200°F to ice at 30°F, you need to remove 144 Btu from one pound of water. This process involves cooling the water to its freezing point of 32°F, then removing the latent heat of fusion to convert it to ice at 32°F, and finally further cooling the ice to the desired temperature of 30°F.
Let f = 0.8888...Then 10*f = 8.8888...Subtracting the first from the second gives 9*f = 8So that f = 8/9.
-200 degrees Celsius is equal to -328 degrees Fahrenheit.
200 F = (200 - 32) x 5/9 Cesius
Let f = 0.161616...Then 100*f = 16.161616...Subtracting the first from the second gives 99*f = 16So that f = 16/99.