Question

what is 25.0 g sample of warm water at 40.0⁰c was added to a 25.0 g sample of water in a styrofoam coffee cup calorimeter initially at 20.0⁰c. the final temperature of the mixed water and calorimeter was 29.5⁰c. calculate the heat capacity of the coffee cup calorimeter. the specific heat of water is 4.184 j/g∙⁰c?

Answer 1

The heat capacity of the coffee cup calorimeter is 0 J/°C.

Step-by-step explanation:

To calculate the heat capacity of the coffee cup calorimeter, we can use the equation:

q = mcΔT

Where q is the heat gained or lost, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

In this case, the heat gained or lost is equal to 0 because the coffee cup calorimeter does not exchange heat with its surroundings. Therefore, we can rearrange the equation to solve for the heat capacity:

q = CΔT

Where C is the heat capacity of the calorimeter.

Plugging in the given values, we have:

C = q/ΔT = 0/9.5 = 0 J/°C