Any help would be great, even if it is just a way to start or a formula.

Calculate the heat necessary to convert 10.0 g of water (just melted) at 0 degrees Celcius to water at 20 degrees Celcius, assuming that the specific heat remains constant at 1 cal/gram degrees celcius.

i used the formula m(delta T)Cp=delta H

(10.0)(20.0-0.00)= x

and so x = 200... is that correct, if so would the units be degrees celcius?

(10.0 g water) (delta 20 degrees C)(4.184J/g x C) = 836.8 J needed to heat the water.

You just forgot to multiply the specific heat.

To calculate the heat necessary to convert 10.0 g of water from 0 degrees Celsius to 20 degrees Celsius, you can use the formula:

Q = m * C * ΔT

Where:
Q = heat energy (in calories)
m = mass of the substance (in grams)
C = specific heat capacity (in cal/gram degrees Celsius)
ΔT = change in temperature (in degrees Celsius)

In this case, the mass (m) is given as 10.0 g and the specific heat (C) is given as 1 cal/gram degree Celsius. To find the change in temperature (ΔT), subtract the initial temperature from the final temperature:

ΔT = final temperature - initial temperature
= 20 degrees Celsius - 0 degrees Celsius
= 20 degrees Celsius

Now, you can plug the values into the formula to calculate the heat (Q):

Q = 10.0 g * 1 cal/g degrees Celsius * 20 degrees Celsius
= 200 cal

Therefore, the heat necessary to convert 10.0 g of water from 0 degrees Celsius to 20 degrees Celsius is 200 calories.