A piece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °C and dropped into 84 g of water (with an initial temperature of 20 °C) in a calorimeter. The final temperature of the system is 52.1°C. The specific heat of water is 4.184 J/g*⁰C. What is the specific heat of the metal?
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T (water) = T (metal) = 52.1 °c
t(w) = 20°c
t(m) = 100°c
hence
Q(w) = mc(T-t)= heat transfer to water
Q(w) = 84×4.184(52.1-20)
= 11281.74 J
since
Q(w) = -Q(m)
=> 11281.74 = -{68.6×C×(52.1-100)}
=> 11281.74/3285.94 = C(metal)
hence
C(m) = 3.43 J/g.°c
t(w) = 20°c
t(m) = 100°c
hence
Q(w) = mc(T-t)= heat transfer to water
Q(w) = 84×4.184(52.1-20)
= 11281.74 J
since
Q(w) = -Q(m)
=> 11281.74 = -{68.6×C×(52.1-100)}
=> 11281.74/3285.94 = C(metal)
hence
C(m) = 3.43 J/g.°c
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