How much energy is needed to raise the temperature of 87g of water from 0*F to 212*F.?

Question

Express your anwser in calories.

Answer 1

Since your answer needs to be in metric units (and the mass given already is) the first thing to do is to convert the temperatures to Celsius.

Deg.C = 5/9 (Deg.F - 32)
0 F = -17.8 C
212 F = 100 C

Then, since the water will go from solid to liquid at some point in there, you need to know three different values: specific heat of ice, latent heat of fusion of water, and specific heat of liquid water. They are (in calories):
0.5 cal / g-degC
80 cal / g
1.0 cal / g-degC

Step 1: (warming ice)
H1 = 87g * 17.8 degC * (0.5 cal / g-degC)
H1 = 774.3 cal

Step 2: (melting ice)
H2 = 87g * (80 cal / g)
H2 = 6960 cal

Step 3: (warming water)
H3 = 87g * 100 degC * (1.0 cal / g-degC)
H3 = 8700 cal

Step 4: add them all up
H1 + H2 + H3 = 774.3 cal + 6960 cal + 8700 cal
Total heat = 16,434.3 cal

Answer 2

You can't use Fahrenheit, so convert to Celsius because the specific heat of water is 4.184 J/g'C or 1 cal/g'C. Then use q = m x C x delta T. Make sure you always use final temp - initial temp for delta T.

Answer 3

Q=MCAT
Q=(87g)(4.184J)(212*F-0*F)
MATH LATIDATIDA!!!!
Divide by 4.184 Because of Calorie's to Joules
Problem solved

Answer 4

4.15 is a value of c for water. its called specific heat capacity. you need to convert your temperature from farentheight to kelvin to use this value of c or either look up a value of c that lets you use farenheit. mutiply your grams x c x temp and you will get an answer in joules. go back to the referce section again to find a conversion formula to go from energy in joule to energy in calories.