The same amount of heat entering identical masses of different substances produces different temperature changes. Calculate the final temperature when 1.25 kcal of heat enters 1.25 kg of the following, originally at 20.0°C.

Question

(a) water
(b) concrete
(c) steel
(d) mercury

Answer 1

1.25 kcal = 5233.5 J

c(merc) = 139 J/kg•K
c(steel) = 470 J/kg•K
c(concrete) = 880 J/kg•K
c(water) = 4183 J/kg•K

ΔT=Q/mc =5233.5/1.25• 139 =30.1º => 50,1º
ΔT=Q/mc =5233.5/1.25•470 =8.91º =>
28,91º
ΔT=Q/mc =5233.5/1.25•880 =4.8º =>
24,8º
ΔT=Q/mc =5233.5/1.25•4183 =1º =>
21º