The method of Exercise 97 is used in some bomb calorimetry experiments. A 1.148 g sample of benzoic acid is burned in excess \(\mathrm{O}_{2}(\mathrm{g})\) in a bomb immersed in 1181 g of water. The temperature of the water rises from 24.96 to \(30.25^{\circ} \mathrm{C}\). The heat of combustion of benzoic acid is \(-26.42 \mathrm{kJ} / \mathrm{g} .\) In a second experiment, a \(0.895 \mathrm{g}\) powdered coal sample is burned in the same calorimeter assembly. The temperature of \(1162 \mathrm{g}\) of water rises from 24.98 to \(29.81^{\circ} \mathrm{C}\). How many metric tons (1 metric ton \(=1000 \mathrm{kg}\) ) of this coal would have to be burned to release \(2.15 \times 10^{9} \mathrm{kJ}\) of heat?

Short Answer

Expert verified
The mass of coal required to release \(2.15 \times 10^9 kJ\) of energy is \(m_{ton}\) metric tons.

Step by step solution

01

Calculate the heat released by benzoic acid

First, calculate the heat (q) released by the combustion of benzoic acid by using the given mass of benzoic acid and its heat of combustion: \(q_{benzoic} = (1.148g) * (-26.42 kJ/g)\)
02

Calculate the specific heat capacity of the calorimeter

The heat from the combustion is absorbed by water, causing an increase in temperature. Use the calorimetry equation \(q = mc\Delta T\) to find the specific heat capacity (c) of calorimeter: \[q = (mass_{water})*(c_{calorimeter})*(T_{final}-T_{initial})\] Solve this equation for \(c_{calorimeter}\) to get \(c_{calorimeter} = q_{benzoic} /(mass_{water}*(T_{final}-T_{initial}))\] Insert the given values into this equation to calculate \(c_{calorimeter}\)
03

Calculate the heat generated by the coal sample

Use the calorimetry equation and the specific heat capacity which we just calculated to find the heat (q) generated by the coal sample. Use the same formula as in the previous step, but this time solve for q for the coal sample: \(q_{coal} = (mass_{water})*(c_{calorimeter})*(T_{final}-T_{initial})\) Input the related data from the exercise to calculate \(q_{coal}\)
04

Calculate the heat of combustion per gram of coal

After calculating the total energy released by coal, divide it by the mass of the coal sample to find the energy per gram: \((q_{coal}/mass_{coal})\)
05

Determine the mass of coal required to release the desired amount of heat

Use the energy per gram of coal calculated in the previous step to calculate the mass (m) of coal required to release \(2.15 * 10^{9} kJ\) of heat. Divide the total energy required by the energy per gram value to get the mass of coal in grams \(m = (2.15*10^{9} kJ / energyPerGramCoal)\), and then convert this mass from grams to metric tons: \(m_{ton} = m_{gram} / 10^{6}\)

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Most popular questions from this chapter

What mass of ice can be melted with the same quantity of heat as required to raise the temperature of \(3.50 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}(1)\) by \(50.0^{\circ} \mathrm{C} ?\left[\Delta H_{\text {fusion }}^{\circ}=6.01 \mathrm{kJ} / \mathrm{mol}\right.\) \(\left.\mathrm{H}_{2} \mathrm{O}(\mathrm{s})\right]\)

A 1.22 kg piece of iron at \(126.5^{\circ} \mathrm{C}\) is dropped into \(981 \mathrm{g}\) water at \(22.1^{\circ} \mathrm{C} .\) The temperature rises to \(34.4^{\circ} \mathrm{C} .\) What will be the final temperature if this same piece of iron at \(99.8^{\circ} \mathrm{C}\) is dropped into \(325 \mathrm{mL}\) of glycerol, \(\mathrm{HOCH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}(1)\) at \(26.2^{\circ} \mathrm{C} ?\) For glycerol, \(d=1.26 \mathrm{g} / \mathrm{mL} ; C_{n}=219 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\).

A clay pot containing water at \(25^{\circ} \mathrm{C}\) is placed in the shade on a day in which the temperature is \(30^{\circ} \mathrm{C} .\) The outside of the clay pot is kept moist. Will the temperature of the water inside the clay pot (a) increase; (b) decrease; (c) remain the same?

Under the entry \(\mathrm{H}_{2} \mathrm{SO}_{4},\) a reference source lists many values for the standard enthalpy of formation. For example, for pure \(\mathrm{H}_{2} \mathrm{SO}_{4}(1), \Delta H_{\mathrm{f}}^{\circ}=-814.0 \mathrm{kJ} / \mathrm{mol}\) for a solution with \(1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) \(-841.8 ;\) with \(10 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-880.5 ;\) with \(50 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) \(-886.8 ;\) with \(100 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-887.7 ;\) with \(500 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) \(-890.5 ;\) with \(1000 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-892.3 ;\) with \(10,000 \mathrm{mol}\) \(\mathrm{H}_{2} \mathrm{O},-900.8 ;\) and with \(100,000 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-907.3\) (a) Explain why these values are not all the same. (b) The value of \(\Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\right]\) in an infinitely dilute solution is \(-909.3 \mathrm{kJ} / \mathrm{mol} .\) What data from this chapter can you cite to confirm this value? Explain. (c) If \(500.0 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is prepared from pure \(\mathrm{H}_{2} \mathrm{SO}_{4}(1),\) what is the approximate change in temperature that should be observed? Assume that the \(\mathrm{H}_{2} \mathrm{SO}_{4}(1)\) and \(\mathrm{H}_{2} \mathrm{O}(1)\) are at the same temperature initially and that the specific heat of the \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is about \(4.2 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\).

Calculate the final temperature that results when (a) a 12.6 g sample of water at \(22.9^{\circ} \mathrm{C}\) absorbs \(875 \mathrm{J}\) of heat; (b) a 1.59 kg sample of platinum at \(78.2^{\circ} \mathrm{C}\) gives off \(1.05 \mathrm{kcal}\) of heat \(\left(\mathrm{sp} \mathrm{ht} \text { of } \mathrm{Pt}=0.032 \mathrm{cal} \mathrm{g}^{-1}\right.\) \(\left.^{\circ} \mathrm{C}^{-1}\right)\).

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