Energy and Specific Heat Report Sheet B. Measuring the Caloric Value of a Food T

Question

Energy and Specific Heat Report Sheet B. Measuring the Caloric Value of a Food Type of Food sample B.1 Mass of aluminum can and water Mass of aluminum can Mass of food Final temperature of water Initial temperature of water B.2 B.3 B.4 Mass of food remaining Calculations B.5 Mass of water B.6 Temperature change for water B.7 Heat gain (calories) of water Show calculations) Heat gain (kilocalories) of water Heat loss (calories) of food sample Heat loss (kilocalories) of food sample B.8 Mass of food undergoing combustion B.9 Caloric value (kcalg) of food sample Show calculations Questions and Problems Q6 A 0.25 g sample of a pretzel is burned. The heat it gives off is used to heat 50. g of water from 18°C to 42°C. What is the caloric value of the pretzel in kcal/g? 40

Explanation / Answer

Ans. Few crucial data, like mass of food, initial and final temperature of water, are NOT mentioned in question.

So, a hypothetical value is used wherever needed for presenting the stepwise solution. Following the steps, you’d be able to do the rest calculations on your own.

#B.2. Mass of food burnt = 1.35 g

#B.5. Mass of water = (Mass of can + water) – Mass of empty can

                                    = 142.83 g – 52.51 g

                                    = 90.32 g

#B.6. Increase in temperature of water, dT = 20.980C

#B.7. Heat gained by a water is given by-

q = m s dT                            - equation 1

Where,

q = heat gained

m = mass

s = specific heat

dT = Final temperature – Initial temperature

Putting the values in equation 1-

q = 90.32 g x (4.184 J g-10C-1) x 20.980C

Or, q = 7928.3185024 J                                         ; [1 cal = 4.184 J]

Or, q = 1894.9136 cal                                            ; [1 kcal = 1000 cal]

            Or, q = 1.895 kilocalorie

# The total amount of heat gained by the water to increase its temperature must be equal to the total amount of heat released during the combustion of food.

So,

            Heat loss of food sample = 1894.9136 cal        

Heat loss of food sample = 1.895 kilocalorie

#B.8. Mass of food undergoing combustion = 1.35 g (assuming complete combustion)

#B.9. Calorific value of food = Energy released / Mass of food burnt

                                                = 1.895 kcal / 1.35 g

                                                = 1.404 kcal/ g

#6. Using equation 1, and “heat gained by water must be equal to the amount of heat released during combustion of petrol”, the amount of heat relasd during combustion fo petrol is given by-

            q = 50.0 g x (1 cal g-10C-1) x (42.0 – 18.0)0C = 1200.0 cal

            Hence, q = 1.20 kcal

Now,

Calorific value of petrol = q / mass of petrol = 1.20 kcal / 0.25 g = 4.8 kcal/g

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