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
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