Question

? Question

Answer 1

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

Answer 2

Answer:57.56°C

Explanation:see attached photo