Question

13.7 kJ of heat energy is transferred to a 2.5 kg object. The temperature of
the object increases from 25°C to 45°C. Determine the specific heat
capacity of the object

Answer 1

Answer:

Calculate the heat actually evolved.

                                            q = mcΔt 

 

Fill in the missing info. We have mL's and we need grams. 

 Use density. (50 mL + 50 mL ) = 100 mL of solution. 

 100 mL X 1     g        =  100 grams of solution. (m = V X D) 

                      mL 

 Find the temperature change. 

        Δt =tfinal - tinitial = 32.4oC - 25.5oC = 6.9oC 

     q = mcΔt 

       = 100 grams X 4.184    J   X 6.9oC 

                                          goC 

        = 2.9 X 103 J 

 This is the heat gained by the water, but in fact it is the heat lost by the reacting HCl and NaOH, therefore q = -2.9 x 103 J.

Hope this answer help you.