The temperature of a sample of water in the liquid phase is changed from 15°C to 25°C by the addition of 810 calories. What is the mass of the water?

Question

The partial answer is: 1X g

Answer 1

q = mC∆T

q = heat = 810 cal

m = mass of water = ?

C = specific heat of water = 1 cal/g/deg

∆T = change in temperature = 25 - 15 = 10 degrees

810 cal = (m)(1 cal/g/deg)(10 deg)

m = 810cal/10 cal/g

m = 81g (not sure why your partial answer is 1X g. Doesn't seem right to me)

Answer 2

We start from the formula:

q = m . s. ΔT

where s is the specific heat capacity of water that equals 1 cal/g.^oC

Solving for m, the mass of water:

m = q/(s.ΔT)

Now, ΔT = Tfinal - Tinitial = (25 - 15)^oC = 10 ^oC

Then:

m = 810 cal/(1 cal/g.^oC . 10^oC) = 81 g