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)

Respuesta :

Answer:

[tex]m_{Co}=6998g=7.0kg[/tex]

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:

[tex]Q_{Co}=-Q_{w}\\\\m_{Co}C_{Co}(T_f-T_{Co})=-m_{w}C_{w}(T_f-T_{w})[/tex]

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

[tex]m_{Co}=\frac{-m_{w}C_{w}(T_f-T_{w})}{C_{Co}(T_f-T_{Co})} \\\\m_{Co}=\frac{-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[/tex]

Best regards!