a sample of cobalt, A, with a mass of 5.00g, is initially at 25 C. When this sample gains 6.70 J of heat, the temperature rises to 27.9 C. Another sample of cobalt, B, with a mass of 7.00 g, iw initially at 25 C. If sample B gains 5.00 J of heat, what is the final temperature of sample B

26.5°C

Explanation:

We can solve this question using the equation:

q = m*S*ΔT

Where Q is heat gained in joules, m is the mass of the sample, S is specific heat and ΔT change in temperature.

With the sample A we can find specific heat of cobalt in order to find, in sample B, the ΔT and the final temperature:

Sample A:

q = m*S*ΔT

6.70J = 5.00g*S*(27.9°C-25.0°C)

0.462J/g°C = Specific heat of cobalt

Sample B:

q = m*S*ΔT

5.00J = 7.00g*0.462J/g°C*ΔT

1.5°C = ΔT

As the initial temperature of sample B is 25°C, final temperature is:

25°C + 1.5°C = 26.5°C

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be