What Is the Combined Thermal Capacitance of the Beaker and Water?

[bertie]

Homework Statement

The mass of the water m_w in a beaker is 0.03988kg
the mass m_g of the glass the beaker is made of is 0.031kg
the specific heat c_p_g of the glass is 800 J/(kg*degrees C)
the specific heat c_p_w of the water is 4178 J/(kg*degrees C)
m_w*c_p_w = C_w where C_w is the thermal capacitance of the water
m_g*c_p_g = C_g where C_g is the thermal capacitance of the glass


What is the combined (lumped, if you will) thermal capacitance of the beaker and the water together?

Homework Equations

m_w*c_p_w = C_w where C_w is the thermal capacitance of the water
m_g*c_p_g = C_g where C_g is the thermal capacitance of the glass

The Attempt at a Solution

Would it be C_combined^-1 = C_w^-1+C_g^-1 the way we would combine 2 electrical capacitors in series?

Thank you all!

 

[mark726]

I can confirm that your approach is correct. The thermal capacitance of the beaker and water together can be calculated by adding the reciprocal of the individual thermal capacitances. In other words, C_combined^-1 = C_w^-1 + C_g^-1. This is similar to how we combine electrical capacitors in series.

However, I would like to point out that the units for thermal capacitance are not in J/(kg*degrees C). The correct units for thermal capacitance are J/(kg*K), where K represents Kelvin, the SI unit for temperature. So, you would need to convert the temperature from degrees Celsius to Kelvin before calculating the thermal capacitance.

Also, keep in mind that this approach assumes that the beaker and water are in thermal equilibrium and there is no heat transfer between them. If there is heat transfer, the combined thermal capacitance will be different and would require a more complex calculation.