Question

A 237 -g piece of molybdenum, initially at \(100.0^{\circ} \mathrm{C},\) was dropped into \(244 \mathrm{g}\) of water at \(10.0^{\circ} \mathrm{C} .\) When the system came to thermal equilibrium, the temperature was \(15.3^{\circ} \mathrm{C} .\) What is the specific heat capacity of molybdenum?

Short answer

The specific heat capacity of molybdenum is approximately \(0.276\, \text{J/g°C}\).

Step-by-step solution

Understand the Problem

We have two substances reaching thermal equilibrium: molybdenum and water. We know the initial temperatures, final equilibrium temperature, and mass of both substances. We need to find the specific heat capacity of molybdenum.

Use the Principle of Heat Exchange

When the molybdenum and water reach equilibrium, the heat lost by molybdenum equals the heat gained by the water. This is expressed as: \[ m_{ ext{Mo}}c_{ ext{Mo}}(T_{ ext{final}} - T_{ ext{initial, Mo}}) = -m_{ ext{water}}c_{ ext{water}}(T_{ ext{final}} - T_{ ext{initial, water}}) \]where \( c_{ ext{Mo}} \) is the specific heat capacity of molybdenum, which we need to find.

Substitute Known Values

Substitute the values into the equation:\[237 imes c_{ ext{Mo}} imes (15.3 - 100.0) = -244 imes 4.18 imes (15.3 - 10.0) \]where the specific heat capacity of water is \( c_{ ext{water}} = 4.18 \, \text{J/g°C} \).

Calculate the Heat Transfer of Water

Calculate the heat gained by water:\[ q_{ ext{water}} = -244 imes 4.18 imes (15.3 - 10.0) = -244 imes 4.18 imes 5.3 = -5382.952 \text{ J} \]

Solve for Specific Heat Capacity of Molybdenum

Now, solve for \( c_{ ext{Mo}} \):\[ -237 imes c_{ ext{Mo}} imes 84.7 = -5382.952 \]\[ 237 imes c_{ ext{Mo}} = \frac{5382.952}{84.7} \]\[ c_{ ext{Mo}} = \frac{5382.952}{237 imes 84.7} \approx 0.276 \, \text{J/g°C} \]

Verify and Conclude

Check if the result makes sense; the unit is appropriate and typical for metals. Conclude that the specific heat capacity of molybdenum found follows expected values for metals.

Key concepts

Thermal Equilibrium

When two objects with different temperatures are in contact, they will eventually reach the same temperature if isolated from other influences. This is known as thermal equilibrium. In the exercise, molybdenum and water are mixed until both reach a final temperature of 15.3°C.

This occurs because heat flows from the hotter substance (molybdenum) to the cooler one (water) until they balance out.

• Hot materials lose heat.
• Cold materials gain heat.

When no more heat flows between them, thermal equilibrium is achieved. This process makes it simpler to calculate and analyze the heat flow between the substances involved.

Heat Transfer

Heat transfer is the movement of thermal energy from one object to another. It can occur in three main ways: conduction, convection, and radiation. In the scenario of the exercise, conduction is the primary form of heat transfer.

Conduction occurs when heat moves through materials that are touching each other, like the molybdenum and water. The metal rod's heat transfers to the water until both objects have the same temperature.

The formula for heat transfer used is:\[ q = mc\Delta T \]
Where:

• \( q \) = heat transferred
• \( m \) = mass
• \( c \) = specific heat capacity
• \( \Delta T \) = change in temperature

This equation helps determine how much heat is gained or lost by each substance.

Principle of Heat Exchange

The principle of heat exchange states that the total heat lost by a hot object equals the total heat gained by a cold object when they reach thermal equilibrium.
This is often summarized by the equation:\[ m_{\text{hot}}c_{\text{hot}}(T_{\text{final}} - T_{\text{initial, hot}}) = -m_{\text{cold}}c_{\text{cold}}(T_{\text{final}} - T_{\text{initial, cold}}) \]
In the exercise, this principle helps to find the specific heat capacity of molybdenum.

Applying it, we see:

• Heat lost by molybdenum = Heat gained by water
• Ensures energy conservation in a closed system

This allows us to solve for unknown variables in a system, like the specific heat capacity, using known values for mass, temperature changes, and specific heat of water.