Answer:
1.1°C
Explanation:
Heat lost by bronze is
H = mc(T2-T1)
H is the heat,
m is mass of bronze 62.3 gram
T1 = initial temperature = 72.30°C
T2 = final temperature = 66.30°C
c is heat capacity, 1.047J/g-°C
Therefore,
H = 62.3 x 1.047 x (66.3 - 72.3)
H = -391.37 Joules
The -ve sign indicate that the heat is lost to the neon.
This amount of heat is gained by the neon.
For the neon,
m = 15.40 gram
c = 0.3770 J/g-°C
T1 =?
T2 = 66.3°C
Equating to the heat gained we have
391.37 = 15.40 x 0.377 x (66.3 - T1)
391.37 = 5.8058 x (66.3 - T1)
67.4 = 66.3 - T2
T1 = 67.4 - 66.3 = 1.1°C
