Question

a sample of cobalt (specific heat of Co=0.418J/G C )at 100.0 C is dropped into a calorimeter containing 500.0 mL of water at 21.1 C. the final temperature of the water/cobalt mixture was 67.1 C. what was the mass of the sample of cobalt? ( Hint 1 mL of water has a mass of 1 gram)

Answer 1

`m_(Co)=6998g=7.0kg`

Step-by-step explanation:

Hello there!

In this case, according to this equilibrium temperature problem, we can set up the following equation to relate the mass, specific heat and temperature change:

`Q_(Co)=-Q_(w)\\\\m_(Co)C_(Co)(T_f-T_(Co))=-m_(w)C_(w)(T_f-T_(w))`

Thus, we solve for the mass of cobalt as shown below:

`m_(Co)=(-m_(w)C_(w)(T_f-T_(w)))/(C_(Co)(T_f-T_(Co))) \\\\m_(Co)=(-500.00g*4.184J/g\°C(67.1\°C-21.1\°C))/(0.418J/g\°C(67.1\°C-100\°C)) \\\\m_(Co)=6998g=7.0kg`

Best regards!