Author: impushpshikha

A 110.0 g piece of zinc at 95.0°C is placed into 60.0 g of water inside a 100.0 g calorimeter at 20.0°C. Given the specific heat capacities of zinc (0.377 kJ/kg°C) and the calorimeter (0.419 kJ/kg°C), determine the final temperature of the system once it reaches thermal equilibrium.

Interpretation

This problem is a classic application of calorimetry, which is based on one of the most important ideas in thermodynamics: energy is conserved.

When a hot object and a cold object are brought into thermal contact inside an insulated system, heat does not disappear; it simply flows from the hotter body to the colder one until a common temperature is reached. That common temperature is called the thermal equilibrium temperature.

In this case:

• The zinc sample starts at $95.0^\circ\text{C}$, so it will lose heat.
• The water and the calorimeter start at $20.0^\circ\text{C}$, so they will gain heat.
• No heat is assumed to escape to the surroundings.

Therefore,

$$\text{Heat lost by zinc} = \text{Heat gained by water} + \text{Heat gained by calorimeter}$$

A common mistake is to include only the water. The calorimeter also absorbs heat, so its heat capacity must be included in the energy balance.

Concept Application

The amount of heat exchanged by a substance is given by$$q = mc\Delta T$$where:

• $m$ = mass
• $c$ = specific heat capacity
• $\Delta T$ = temperature change

Notice that zinc cools while water and the calorimeter warm. Rather than worrying about signs, it is often clearer to write:

$$\text{Heat lost} = \text{Heat gained}$$

and use positive quantities for each.

The specific heats are given in

Since

we may conveniently work with masses in grams.

For water,

For zinc,

For the calorimeter,

Solution

Step 1: Write the heat-balance equation

Heat lost by zinc:$$q_{\text{Zn}} = (110)(0.377)(95-T_f)$$

Heat gained by water:$$q_w = (60)(4.184)(T_f-20)$$

Heat gained by calorimeter:$$q_{\text{cal}} = (100)(0.419)(T_f-20)$$

Applying conservation of energy:(110)(0.377)(95−Tf​)=(60)(4.184)(Tf​−20)+(100)(0.419)(Tf​−20)

Step 2: Simplify the coefficients

For zinc:$$110 \times 0.377 = 41.47$$

For water:$$60 \times 4.184 = 251.04$$

For the calorimeter:$$100 \times 0.419 = 41.90$$

Substituting:$$41.47(95-T_f) = 251.04(T_f-20) + 41.90(T_f-20)$$

Combining the right-hand side:$$41.47(95-T_f) = 292.94(T_f-20)$$

Step 3: Expand both sides

$$3939.65 – 41.47T_f = 292.94T_f – 5858.8$$

Bring all temperature terms to one side:$$3939.65 + 5858.8 = 292.94T_f + 41.47T_f$$$$9798.45 = 334.41T_f$$

Step 4: Solve for the final temperature

$$T_f = \frac{9798.45}{334.41}$$$$T_f = 29.3^\circ\text{C}$$

Final Answer

$$\boxed{T_f = 29.3^\circ\text{C}}$$

Thus, after thermal equilibrium is established, the zinc, the water, and the calorimeter all reach a common temperature of approximately$$\boxed{29.3^\circ\text{C}}.$$

Insight

A powerful way to check your answer is to compare the heat capacities of the objects:$$m c_{\text{Zn}} = (110)(0.377) \approx 41.5\ \text{J}/^\circ\text{C}$$

while$$m c_{\text{water}} + m c_{\text{cal}} = 251.0 + 41.9 \approx 293\ \text{J}/^\circ\text{C}$$

The water–calorimeter system can absorb about seven times more heat per degree than the zinc can release per degree. Therefore, the final temperature should stay much closer to $20^\circ\text{C}$ than to $95^\circ\text{C}$. The calculated value of $29.3^\circ\text{C}$ is exactly what we would expect physically.

Memory aid: In calorimetry, do not focus first on masses—focus on $mc$, the heat capacity. The object with the larger heat capacity has the greater influence on the final equilibrium temperature.

Introduction

Think about traveling from Delhi to Mumbai. You might go by train, bus, or airplane. The journey may differ in route and experience, but the starting point and destination remain the same, so the overall displacement does not change.

A similar idea appears in everyday physics. If you move from the first floor to the fourth floor of a building, you may go directly or stop at intermediate floors. Regardless of the path, the total gain in gravitational potential energy is the same because it depends only on where you start and where you end.

Chemical reactions follow this same principle.

When reactants are converted into products, the total energy change, specifically the enthalpy change depends only on the initial state (reactants)and the final state (products). It does not matter whether the reaction occurs in one step or through several intermediate steps.

This powerful idea is formalized in Hess’s Law, proposed by the Swiss chemist Germain Hess.

Statement of Hess’s Law

Hess’s Law states:

The total enthalpy change \(ΔH\) for a chemical reaction is the same whether the reaction occurs in a single step or in multiple steps.

In other words, if a reaction can be expressed as the sum of several smaller reactions, then:\(ΔHoverall​=ΔH1​+ΔH2​+ΔH3​+⋯\)

This works because enthalpy is a state function.

Understanding the Key Idea: State Function

A state function is a property that depends only on the current state of the system, not on how the system reached that state.

Enthalpy is one such property. Therefore:\(ΔH=Hproducts​−Hreactants​\)

This expression tells us something important: once the initial and final states are fixed, the value of \(ΔH\) is fixed. The path taken in between becomes irrelevant.

What Does Enthalpy Change Represent?

The enthalpy change \(ΔH\) represents the heat absorbed or released at constant pressure.

• If \(ΔH<0\), heat is released (exothermic reaction)
• If \(ΔH>0\), heat is absorbed (endothermic reaction)

Since enthalpy is a state function, this heat change depends only on the initial and final conditions, not on the reaction pathway.

Illustration with a Chemical Reaction

Let us apply Hess’s Law to a real example.

We want to determine the enthalpy change for the formation of carbon monoxide:

Direct measurement of this reaction is not easy. So instead, we use known reactions:

Our goal is to obtain the target reaction.

We notice that in equation (b), CO appears as a reactant, but in our target reaction, CO must be a product.

So we reverse equation (b).

Step 1: Reverse the Reaction

$\text{CO}_2(g)\rightarrow \text{CO}(s) + \frac{1}{2}\text{O}_2(g) \ ,ΔH=+283.0 kJ/mol$

Why did the sign change?

Because reversing a reaction reverses the direction of heat flow. A reaction that originally released heat will now require the same amount of heat.

Step 2: Add the Equations

Now, add equation (a) and the reversed (b):\(C(s)+O2​(g)→CO2​(g)\)

Cancel species that appear on both sides:

• \(CO2​(g)\) cancels out
• One-half of \(O2​(g)\) remains

The final equation becomes:

Step 3: Add Enthalpy Changes\(ΔH=(−393.5)+(283.0)=−110.5 kJ/mol\)

This matches the enthalpy change for the direct formation of CO.

Algebraic Rules for Applying Hess’s Law

Hess’s Law works like algebra, and the rules are logical once you see why they exist:

• Reversing a reaction changes the sign of \(ΔH\), because heat flow reverses.
• Multiplying a reaction multiplies \(ΔH\), since energy is an extensive property.
• Adding reactions means adding their enthalpy changes.
• Common species cancel, just like terms in equations, because they do not affect the net change.

Conclusion

Hess’s Law teaches a powerful and elegant idea:
Energy changes depend only on where a reaction starts and where it ends—not on how it gets there.

This allows chemists to calculate enthalpy changes for reactions that are difficult or impossible to measure directly.

Once you internalize this, Hess’s Law stops being a formula to memorize and becomes a logical tool—almost like solving a puzzle where energy must always balance perfectly.