I guess both the cup and the water started at 30 C and ended at 35 C
so they went up 5 C
heat in = .6*.22*5+2.4*1*5 = 12.66 KCal
heat out = 2.34 * c * (210-30)
= 421 c KCal
heat in = heat out
12.66 = 421 c
c = 0.03
are you sure you added 2.34 kg of this stuff and not maybe 2.34 grams?
An aluminum calorimetry cup [mc = 0.6 kg, cc = 0.22 kcal/(kgC)] is filled with water [mw = 2.4 kg, cw = 1 kcal/(kgC)]. It is at a temperature of 30C. A sample of an unknown substance with mass mm = 2.34 kg, originally at a temperature of 210C, is added. After a while, the temperature of the water is 35C. What is the specific heat of this material?
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