Heat Equation Problem

Question

A 40 gram piece of zinc at 45 celsius is placed in a beaker of cool water. If the zinc loses 375 joules of energy as it is cooled to 21 Celsius, determine the specific heat capacity of zinc.

Answer 1

Hello Anna

 

The equation to use is:

 

c = q / (m ΔT)

 

where c is the specific heat capacity, q is the amount of heat absorbed or released, m is the mass and ΔT is the change in temperature (Tf - Ti). Your question lists the following data:

 

m = 40 g

Ti = 45 ºC

Tf = 21 ºC

q = -375 J (q is negative because heat was lost, hence, exothermic).

 

ΔT = Tf - Ti = 21 - 45 = -24 ºC

 

Now plug in all the numbers and solve for c.

 

c = q / (m ΔT)

 

   = -375 J / (40 g x -24 ºC)

 

   = 0.39 J/g ºC

~ 0.4 J/g ºC(rounded to 1 significant figure)