A sample of cobalt, A, with a mass of 5g is initially at 25 degrees Celsius. When this sample gains 6.70J of heat, the temperature rises to 27.9 degrees Celsius. Another sample of cobalt B with a mass of 7 g is initially at 25 degrees Celsius. If sample B gains 5 J of heat, what is the final temp of B. I

Question

A sample of cobalt, A, with a mass of 5g is initially at 25 degrees Celsius. When this sample gains 6.70J of heat, the temperature rises to 27.9 degrees Celsius. Another sample of cobalt B with a mass of 7 g is initially at 25 degrees Celsius. If sample B gains 5 J of heat, what is the final temp of B. I

Honestly don't know how to do this.

Answer 1

A two-step problem. First, determine the specific heat of Co from sample A.

q = mass Co x specific heat Co x (Tfinal-Tinitial). The only unknown in this equation is specific heat Co in sample A.

Specific heat doesn't change. NHow use the same formula for sample B of Co.
q = mass Co x sp.h. Co x (Tfinal-Tinitial). The only unknown in this equation is Tfinal.

Answer 2

To solve this problem, we need to use the concept of specific heat and the equation:

q = mcΔT

where:
- q is the amount of heat gained or lost by the substance (in joules, J)
- m is the mass of the substance (in grams, g)
- c is the specific heat capacity of the substance (in J/g°C)
- ΔT is the change in temperature (in °C)

In this case, we are given the amount of heat gained and the initial temperature for sample A. Using this information, we can calculate the specific heat capacity of cobalt and then use it to find the final temperature of sample B.

Let's start by calculating the specific heat capacity of cobalt (c).

First, rearrange the equation to solve for c:

c = q / (m * ΔT)

Using the values given for sample A:
q = 6.70 J
m = 5 g
ΔT = 27.9°C - 25°C = 2.9°C

c = 6.70 J / (5 g * 2.9°C)
c ≈ 0.460 J/g°C

Now that we have the specific heat capacity for cobalt, we can use it to find the final temperature of sample B.

Using the same equation:

q = mcΔT

Rearranging the equation to solve for ΔT:

ΔT = q / (mc)

Using the values for sample B:
q = 5 J
m = 7 g
c ≈ 0.460 J/g°C

ΔT = 5 J / (7 g * 0.460 J/g°C)
ΔT ≈ 1.087°C

To find the final temperature of sample B, we add the change in temperature to the initial temperature of 25°C:

Final temperature of B = 25°C + ΔT ≈ 25°C + 1.087°C ≈ 26.09°C

Therefore, the final temperature of sample B is approximately 26.09°C.