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Chapter 5: Question 16 E (page 269)

Would the amount of heat absorbed by the dissolution in example 5.6 appear greater, lesser, or remain the same if the experimenter used a calorimeter that was a poorer insulator than a coffee cup calorimeter? Explain your answer.

Short Answer

Expert verified

The amount of heat measured would be lesser if the calorimeter was replaced with a poorly insulated calorimeter.

Step by step solution

01

Change in temperature

The amount of heat measured in example 5.6 is 1kJ.

If the calorimeter is poorly insulated, it will lose heat to its surroundings, decreasing the overall temperature of the reaction. Therefore, the change in temperature would also decrease.

02

Heat measured

The heat required to raise the temperature of a substance is given by the formula Q = C \( \times \)m\( \times \)\(\Delta \)T, where 鈥淐鈥� is the specific heat of the substance, 鈥渕鈥� is the mass of the substance, and 鈥溾垎T鈥� is the change in the temperature of the substance.

Suppose the change in temperature decreases, and the heat measured for the reaction would also decrease. So, the heat measured for the reaction in example 5.6 would be less than1 kJ.

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Most popular questions from this chapter

Aluminum chloride can be formed from its elements:

(i)\({\bf{2Al(s) + 3C}}{{\bf{l}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2AlC}}{{\bf{l}}_{\bf{3}}}{\bf{(s) \Delta H^\circ = ?}}\)

Use the reactions here to determine the 螖H掳 for reaction(i):

\(\begin{array}{*{20}{l}}{\left( {{\bf{ii}}} \right){\rm{ }}{\bf{HCl(g)}} \to {\bf{HCl(aq) \Delta H^\circ (ii) = - 74}}{\bf{.8 kJ}}}\\{\left( {{\bf{iii}}} \right){\rm{ }}{{\bf{H}}_{\bf{2}}}{\bf{(g) + C}}{{\bf{l}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2HCl(g) \Delta H^\circ (iii) = - 185 kJ}}}\\{\left( {{\bf{iv}}} \right){\rm{ }}{\bf{AlC}}{{\bf{l}}_{\bf{3}}}{\bf{(aq)}} \to {\bf{AlC}}{{\bf{l}}_{\bf{3}}}{\bf{(s) \Delta H^\circ (iv) = + 323 kJ}}}\\{\left( {\bf{v}} \right){\rm{ }}{\bf{2Al(s) + 6HCl(aq)}} \to {\bf{2AlC}}{{\bf{l}}_{\bf{3}}}{\bf{(aq) + 3}}{{\bf{H}}_{\bf{2}}}{\bf{(g) \Delta H^\circ (v) = - 1049 kJ}}}\end{array}\)

Calculate \({\bf{\Delta H}}\)for the process Hg2Cl2(s)鉄�2贬驳(l) + Cl2(g)

from the following information:

Hg(l) + Cl2(g)鉄禜驳颁濒2(s) 螖H= 鈭�224 kJ

Hg(l) + HgCl2(s)鉄禜驳2Cl2(s) 螖H= 鈭�41.2 kJ

Question: Calculate the heat capacity, in joules and in calories per degree, of the following:

(a) 28.4 g of water

(b) 1.00 oz of lead

When 100 mL of 0.200 M NaCl(aq) and 100 mL of 0.200 M AgNO3(aq), both at 21.9 掳C, are mixed in a coffee cup calorimeter, the temperature increases to 23.5 掳C as solid AgCl forms. How much heat is produced by this precipitation reaction? What assumptions did you make to determine your value?

One method of generating electricity is by burning coal to heat water, which produces steam that drives an electric generator. To determine the rate at which coal is to be fed into the burner in this type of plant, the heat of combustion per ton of coal must be determined using a bomb calorimeter. When 1.00g of coal is burned in a bomb calorimeter (figure 5.17), the temperature increases by 1.48藲C. If the heat capacity of the calorimeter is 21.6 kJ/藲C, determine the heat produced by the combustion of a ton of coal (2.000 脳 103).
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