Question

If 0.507J of heat leads to a 0.007 degree C change in water, what mass is present?

Answer 1

17.32 gm

Step-by-step explanation:

Specific heat of water is 4.182 J / g- c

4.182 J /g-c = .507/ ( .007 * gm)

solve for gm = 17.32 gm

Answer 2

17.25g

Step-by-step explanation:

From our question, we have been asked to find the mass of water present.

The formula for finding specific heat is given by;

Q = mCp∆T

We have, Q = 0.507 J, ∆T = 0.007° C and we know that Cp of water is 4.2 J/g°C

Rearrange the formula making m the subject.

Which is given by;

m = (Q)/(Cp∆T)

Substituting for the known values

`m \: = (0.507)/(4.2 * 0.007)`

= 17.24489 g

= 17.25g