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

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2