College Chemistry 1 Question

Question

When a 45 g sample of an alloy at 100ºC is dropped into 100.0 g of water at 25.0ºC the final temperature is 37.0ºC. What is the specific heat of the alloy?

Answer 1

heat gained by water = q = mC∆T

m = mass = 100.0 g

C = specific heat of water = 4.184 J/g/deg

∆T = change in temperature = 37 - 25 = 12 degrees

q = 100.0g x 4.184 J/g/deg x 12 deg = 5021 J

This must also be the heat lost by the alloy, and the temperature change of the alloy is100 - 37 = 63 deg

Specific heat is joules/g/deg = 5021 J/45 g /63 deg = C = 1.77 J/g/deg

Done another way:

heat lost by alloy = heat gained by water

(45.0 g)(C J/g/deg)(63 deg) = (100.0 g)(4.184 J/g/deg)(12 deg)

2835 C = 5021

C = 1.77 J/g/deg