Asked by Eric

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

Answers

Answered by Elena
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

Related Questions