Example 1

A 360 g piece of rebar (a steel rod used for reinforcing concrete) is dropped into 425 mL of water at 24.0 °C. The final temperature of the water was measured as 42.7 °C. Calculate the initial temperature of the piece of rebar. Assume the specific heat of steel is approximately the same as that for iron (Specific Heat Table), and that all heat transfer occurs between the rebar and the water (there is no heat exchange with the surroundings).

Solution

The temperature of the water increases from 24.0 °C to 42.7 °C, so the water absorbs heat. That heat came from the piece of rebar, which initially was at a higher temperature. Assuming that all heat transfer was between the rebar and the water, with no heat “lost” elsewhere, then heat given off by rebar = −heat taken in by water, or:

q_rebar = –q_water

Since we know how heat is related to other measurable quantities, we have:

(c × m × ∆T)_rebar = -(c × m × ∆T)_water

Letting f = final and i = initial, in expanded form, this becomes:

c_rebar × m_rebar × (T_{f, rebar} – T_{i, rebar}) = -c_water × m_water × (T_{f, water}– T_{i, water})

The density of water is 1.0 g/mL, so 425 mL of water = 425 g. Noting that the final temperature of both the rebar and water is 42.7 °C, substituting known values yields:

(0.449 J/(g·°C))(360 g)(42.7 °C – T_{i, rebar}) = (4.184 J/(g·°C))(425 g)(42.7 °C – 24.0 °C)

Solving this gives T_{i, rebar} = 248 °C, so the initial temperature of the rebar was 248 °C.

Check Your Learning

A 248 g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C. Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water.

Example 2

A 59.7 g piece of metal that had been submerged in boiling water was quickly transferred into 60.0 mL of water initially at 22.0 °C. The final temperature is 28.5 °C. Use these data to determine the specific heat of the metal. Use this result to identify the metal.

Solution

Assuming perfect heat transfer, heat given off by metal = −heat taken in by water, or:

q_metal + q_water = 0

In expanded form, this is:

c_metal × m_metal × (T_{f, metal} – T_{i, metal}) = -c_water × m_water × (T_{f, water} – T_{i, water})

Noting that since the metal was submerged in boiling water, its initial temperature was 100.0 °C; and that for water, 60.0 mL = 60.0 g; we have:

(c_metal)(59.7 g)(28.5 °C – 100.0 °C) = -(4.184 J/(g·°C))(60.0 g)(28.5 °C – 22.0 °C)

Solving for c_metal gives:

Comparing this with values in the Specific Heats Table, our experimental specific heat is closest to the value for copper (0.39 J/(g·°C)), so we identify the metal as copper.

Check Your Learning

A 92.9 g piece of a silver/gray metal is heated to 178.0 °C, and then quickly transferred into 75.0 mL of water initially at 24.0 °C. After 5 minutes, both the metal and the water have reached the same temperature: 29.7 °C. Determine the specific heat and the identity of the metal. (Note: You should find that the specific heat is close to that of two different metals. Explain how you can confidently determine the identity of the metal).

Example 3

When 50.0 mL of 0.10 M HCl(aq) and 50.0 mL of 0.10 M NaOH(aq), both at 22.0 °C, are added to a calorimeter, the temperature of the mixture reaches a maximum of 28.9 °C. What is the approximate amount of heat produced by this reaction?

HCl(aq) + NaOH(aq)  →  NaCl(aq) + H₂O(ℓ)

Solution

To visualize what is going on, imagine that you could combine the two solutions so quickly that no reaction took place while they mixed; then after mixing, the reaction took place. At the instant of mixing, you have 100.0 mL of a mixture of HCl and NaOH at 22.0 °C. The HCl and NaOH then react until the solution temperature reaches 28.9 °C.

The heat given off by the reaction is equal to that absorbed by the solution. Therefore:

q_reaction = –q_solution

(It is important to remember that this relationship only holds if the calorimeter does not absorb any heat from the reaction, and there is no heat exchange the surroundings outside of the calorimeter.)

Next, we know that the heat absorbed by the solution depends on its specific heat, mass, and temperature change:

q_solution = (c × m × ΔT)_solution

To proceed with this calculation, we need to make a few more reasonable assumptions or approximations. Since the solution is aqueous, we can proceed as if it were water in terms of its specific heat and mass values. The density of water is approximately 1.0 g/mL, so 100.0 mL has a mass of about 1.0 × 10² g (two significant figures). The specific heat of water is approximately 4.18 J/g °C, so we use that for the specific heat of the solution. Substituting these values gives:

q_solution = (4.184 J/(g·°C))(1.0 × 10² g)(28.9 °C – 22.0 °C) = 2.89 × 10³ J

Finally, since we are trying to find the heat of the reaction, we have:

q_reaction = –q_solution = -2.89 × 10³ J

The negative sign indicates that the reaction is exothermic. It produces 2.89 kJ of heat.

Check Your Learning

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

Chemistry in Real Life: Thermochemistry of Hand Warmers

When working or playing outdoors on a cold day, you might use a hand warmer to warm your hands. A common reusable hand warmer contains a supersaturated solution of NaC₂H₃O₂ (sodium acetate) and a metal disc. Bending the disk creates nucleation sites around which the NaC₂H₃O₂ quickly crystallizes (a later chapter on solutions will investigate saturation and supersaturation in more detail).

The process

NaC₂H₃O₂(aq)  →  NaC₂H₃O₂(s)

is exothermic, and the heat produced by this process is absorbed by your hands, thereby warming them (at least for a while). If the hand warmer is reheated, the NaC₂H₃O₂ redissolves and can be reused.

Another common hand warmer produces heat when it is ripped open, exposing iron and water in the hand warmer to oxygen in the air. One simplified version of this exothermic reaction is

2Fe(s) + O₂(g)  →  Fe₂O₃(s)

Salt in the hand warmer catalyzes the reaction, so it produces heat more rapidly; cellulose, vermiculite, and activated carbon help distribute the heat evenly. Other types of hand warmers use lighter fluid (a platinum catalyst helps lighter fluid oxidize exothermically), charcoal (charcoal oxidizes in a special case), or electrical units that produce heat by passing an electrical current from a battery through resistive wires.

Example 4

When solid ammonium nitrate dissolves in water, the solution becomes cold. This is the basis for an “instant ice pack” (Figure 3). When 3.21 g of solid NH₄NO₃ dissolves in 50.0 g of water at 24.9 °C in a calorimeter, the temperature of the solution decreases to 20.3 °C.

Calculate the value of q for this reaction and explain the meaning of its arithmetic sign. State any assumptions that you made.

Solution

We assume that the calorimeter prevents heat transfer between the solution and its external environment (including the calorimeter itself), in which case:

q_reaction = –q_solution

Assuming also that the specific heat of the solution is the same as that for water, we have:

q_reaction = –q_solution = -(c × m × ΔT)_solution
q_reaction = -[(4.184 J/(g·°C)) × (53.2 g) × (20.3 °C – 24.9 °C)]
= -[(4.184 J/(g·°C)) × (53.2 g) × (-4.6 °C)]
= +1.0 × 10³ J = +1.0 kJ

The positive sign for q_reaction indicates that the dissolution is an endothermic process.

Check Your Learning

When a 3.00-g sample of KCl was added to 3.00 × 10² g of water in a calorimeter, the temperature decreased by 1.05 °C. How much heat is involved in the dissolution of the KCl? What assumptions did you make?