How do you use the equation for heat?

Question

How do you solve this problem with the equation:
Q = mc (delta) T
What amount of heat is required to heat 5.00 grams of liquid water from 0.00 degrees C to 100.0 degrees C?

Answer 1

Q is the quantity of heat (in ergs, Joules etc.) which is needed to increase a mass m of a substance with a specific heat c (sometimes we use s also) through a temperature differential delta T.

For water, specific heat is 1. So, fr 5 grams of water to be heated from 0 to 100 C = 5.1.100 = 500 ergs.

Obseve that we are dealing only with liquid water. IF we had 5 grms of ice at 0 C and we needed 5 grams of steam at 100 C, we would have needed to add the latent heat of melting of ice and latent heat of boiling of water.

Answer 2

The question is a straight forward usage of the formula.
m = mass = 5.00 g
c = specific heat 1.0cal/g-C or 4.184 J/g-K
delta T = T2 -T1 = 100.0 degrees C or Kelvins

Q = 5.00g x 1.00 cal/g-C x 100 C= 500 cal
Q = 5.00 g x 4.184 J/g-K x 100 K = 2092 J = 2090 J (sig fig)

Answer 3

q=heat
m=mass
c=specific heat of substance, of water 1 J/(g x C)
/\(delta)T=Final temp. - initial temp.

q=5.00g x 1 J/(g x C) x (100C-0C)