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
(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
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