Question

how many grams of granite at 54.1 c must be added to 467.62 of water at 21 C to make the final temp of both come out to be 29.2 C? Specific heat of granite 0.79 J/g C, specific heat of water 4.184 J/g C

Answer 1

815.59

Step-by-step explanation:

Assuming no heat lost to the Surrounding

Total heat lost by granite = Total heat gained by water

mc∆T of granite = mc∆T of water

m x 0.79 x (54.1 - 29.2) = 467.62 x 4.184 x (29.2 - 21.0)

m x 19.671 = 16043.48106

m = 16043.48106/19.671

= 815.59J