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

heat acquired by water = q = mC∆T

m = mass = 100.0 g

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

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

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

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

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

heat lost by alloy = heat acquired 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

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