You need to boil 307 g of water (specific heat = 4.184) so you can make a cup of your favorite drink. If room temperature is 20.15 oC, how much heat is needed to boil the water?

I suppose you mean 4.184 J/g*C for sp.h. H2O.

q = mass H2O x sp.h. x (Tfinal-Tinitial)
q = 307 x 4.184 x (100-20.15).
That will give you q in joules needed to raise the temperature of the water from 20.15 to 100. I will not boil it. You will need an additional 2260 J for each grams of water actually boiled but I don't think the problem is asking for that.

To determine how much heat is needed to boil the water, we can use the formula:

Q = m * c * ΔT

where:
Q = amount of heat energy
m = mass of the substance (water) in grams
c = specific heat capacity of the substance
ΔT = change in temperature

Given:
m = 307 g
c = 4.184 J/g°C (specific heat of water)
ΔT = the change in temperature from room temperature (20.15 °C) to boiling point (100 °C)

To calculate ΔT, we subtract the initial temperature from the final temperature:

ΔT = 100 °C - 20.15 °C
= 79.85 °C

Now we can substitute the values into the formula to find Q:

Q = (307 g) * (4.184 J/g°C) * (79.85 °C)
≈ 100,578.4 J

Therefore, approximately 100,578.4 J of heat energy is needed to boil 307 g of water.