High School Chem problem

Question

A student is given the task to determine the mass of a piece of rebar (a steel rod used for reinforcing concrete) using the method of simple calorimetry. The specific heat of steel is 0.449 J/g°C. The piece of rebar is initially heated to 248.3°C and then dropped into the calorimeter which contained 425 mL of water, which has a specific heat of 4.184 J/g°C, at 24.0°C. The final temperature of the mixture was measured as 42.7°C.  

(a) Write a thermochemical equality to display the heat transfer that would occur in the calorimeter.

(b) Calculate the mass of the rebar that was used in the experiment.

                 

(c) Identify a significant error that commonly occurs during simple calorimetry experiments. Explain the impact this would have on the students calculated mass of the rebar.

(d) If the experiment were repeated with an equal mass of aluminum in place of the rebar, explain what would happen to the final temperature measured in the calorimeter. The known specific heat of aluminum is 0.897 J/g°C. Assume an equal volume of water is used in the calorimeter, and the initial temperature of the water and aluminum are the same as in the original experiment.

Answer 1

(a) heat lost by steel = heat gained by water

-q_steel = +q_H2O

(b) heat lost by steel = - (mass)(specific heat)(change in temp) = (mass)(0.449J/gº)(205.6º) = 92.3 J/g x mass

heat gained by H2O = (mass)(specific heat)(change in temp) = (425 g)(4.184J/gº)(18.7º) = 33,252 J

(92.3 J/g)(mass) = 33,252 J

mass = 360 g

(c) The error most commonly encountered in simple calorimetry is that heat is lost to the surroundings because the calorimeter isn't completely insulated. The net result is that the heat transferred is underestimated making the mass of the steel also be underestimated.

(d) Since the specific heat for Al is about twice that for steel, the final temperature will be greater than it was when steel was used. This is because the definition of specific heat is the amount of heat needed to raise the temperature of 1 g of material by 1 degree. So it takes more heat to raise the temp of Al than it does to raise the temp of steel, and this extra heat will be transferred to the water, so the final temp of the water will be greater than 42.7ºC . This can be verified by the following calculations where Tf represents the final temperature:

(360 g)(0.897 J/gº)(248-Tf) = (425 g)(4.184 J/gº)(Tf - 24)

80,084 - 323Tf = 1778Tf - 42,677

Tf = 58.4º