Heat capacity of the coffee cup calorimeter?

Question

A reaction known to release 1.78kJ of heat took place in a coffee cup calorimeter containing 0.100L of solution where the temp rose by 3.65 deg. C.

 

In a second experiment, when 50.0mL of 0.20M HCl and 50.0mL of 0.20M NaOH were mixed in the same calorimeter, the temp rose 1.26 deg. C.

 

Assume the specific heat of all the solutions is 4.18J/g deg. C.

 

Calculate the heat capacity of the coffee cup calorimeter.

 

Answer 1

Hello Haley

 

You haven't stated the units for heat capacity, but I will list it in both kJ / ºC and J / ºC.

 

For experiment #1,

 

q = -1.78 kJ (sign is negative because heat was released)

ΔT = 3.65 ºC

 

Heat capacity, C = q / ΔT = -1.78 kJ / 3.65 ºC = -0.488 kJ / ºC (-488 J / ºC)

 

 

For experiment #2, you need to first calculate the amount of heat (q) released from the reaction.  Notice that the question states that the temp. "rose", which means the reaction is exothermic.

 

The equation to use is:  q = m c ΔT where q is heat, m is the mass of the solution in units of grams, c is the specific heat, and ΔT is the change in temperature.

 

Since 50.0 mL of HCl is mixed with 50.0 mL of NaOH, the total volume of the solution is 50.0 + 50.0 = 100.0 mL.  We will assume that the solution is mostly water, and then use the density of water (1.00 g / mL) to convert the volume to mass.

 

m = d x V = 1.00 g / mL x 100.0 mL = 100.0 g

 

Now plug in the numbers to solve for q.

 

q = 100.0 g x 4.184 J /g ºC x 1.26 ºC = -527 J (q is negative since the reaction is exothermic!)

 

Finally, solve for the heat capacity.

 

C = q / ΔT = -527 J / 1.26 ºC = -418 J / ºC (-0.418 kJ / ºC)