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
To calculate the BTUs needed to convert one pound of water at 200°F to ice at 30°F, we must consider two stages: first, cooling the water from 200°F to 32°F, and then freezing it to ice at 32°F, and finally cooling the ice to 30°F. Cooling from 200°F to 32°F requires removing about 168 BTUs (using the specific heat of water, which is approximately 1 BTU/lb°F). Freezing the water at 32°F to ice requires removing 80 BTUs (the latent heat of fusion). Finally, cooling the ice from 32°F to 30°F requires an additional 2 BTUs (using the specific heat of ice, which is about 0.5 BTU/lb°F). Adding these together: 168 + 80 + 2 = 250 BTUs must be removed.
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