Question

It takes \(452 \mathrm{~J}\) of heat to raise the temperature of a \(36.8 \mathrm{~g}\) sample of a metal from \(22.9^{\circ} \mathrm{C}\) to \(98.2^{\circ} \mathrm{C}\). What is the heat capacity of the metal?

Short answer

The heat capacity of the metal is approximately \(0.163\, \mathrm{J/g^{\circ}C}\).

Step-by-step solution

Understand the Formula

The formula for heat capacity (specific heat) is given as \( c = \frac{q}{m \Delta T} \), where \( q \) is the heat transferred (in joules), \( m \) is the mass of the substance (in grams), and \( \Delta T \) is the change in temperature (in degrees Celsius).

Calculate the Temperature Change

The change in temperature (\( \Delta T \)) can be calculated by subtracting the initial temperature from the final temperature. \[ \Delta T = 98.2^{\circ} C - 22.9^{\circ} C = 75.3^{\circ} C \].

Plug in the Values

Now substitute the given values into the formula: \( q = 452 \; \mathrm{J} \), \( m = 36.8 \; \mathrm{g} \), and \( \Delta T = 75.3^{\circ} C \). The formula becomes \[ c = \frac{452}{36.8 \times 75.3} \].

Perform the Calculation

First, calculate the denominator: \( 36.8 \times 75.3 = 2770.64 \). Then, divide the heat by the product of mass and temperature change: \[ c = \frac{452}{2770.64} \approx 0.163 \; \mathrm{J/g^{\circ}C} \].

Key concepts

Heat Transfer

When we talk about heat transfer in physics, we're discussing how thermal energy moves from one substance to another. Imagine holding a hot mug of tea. The heat from the mug can move to your cooler hands, causing a sensation of warmth. In the context of calorimetry, which is measuring the heat involved in chemical reactions or physical changes, heat transfer helps us understand how energy flows. Here, 452 Joules of heat energy were transferred to our sample of metal, indicating its absorption capacity.

Heat always moves from a substance at a higher temperature to one at a lower temperature until equilibrium is reached. The principle behind heat transfer is crucial when calculating specific heat, which tells us how much heat per unit mass is required to change the temperature. Here are three common methods of heat transfer:

• Conduction - Direct contact heat flow, like your hand touching a hot pot.
• Convection - Fluid-related heat transfer, such as boiling water where hot water rises and cools as it sinks.
• Radiation - Heat transfer through electromagnetic waves, like sunrays warming the earth.

Temperature Change

Temperature change is all about how much hotter or colder a substance becomes. In our exercise, the metal's temperature rose from \(22.9^{\circ} C\) to \(98.2^{\circ} C\). This is an important part of calculating specific heat capacity because, along with mass and heat, it determines how much energy is needed.

To find the change in temperature, we simply subtract the initial temperature from the final temperature: \(\Delta T = 98.2^{\circ} C - 22.9^{\circ} C = 75.3^{\circ} C\). Temperature change gives us a "roadmap" of how the thermal energy affects the object over the range of temperatures. The greater the temperature change, the more energy is involved in transitioning that substance to its new thermal state.

It's vital to note that a larger temperature change, with the same heat input, means the substance has a lower specific heat capacity. Conversely, smaller temperature changes mean higher specific heat capacity for the same heat input.

Calorimetry

Calorimetry is a fascinating technique that helps measure the change in heat during chemical reactions or physical processes. Think of it as being like a thermal detective, where we use the principles of heat transfer and temperature change to understand the specific heat capacity of substances. By knowing the specific heat capacity, you can predict how different materials respond to heat—a critical aspect for everything ranging from cooking to industrial applications.

In this exercise, we're looking into calorimetry to calculate the specific heat (or heat capacity) of a metal. We use the formula \(c = \frac{q}{m \Delta T}\), which requires knowing the heat supplied \(q\), the mass \(m\), and the temperature change \(\Delta T\). This formula helps us discover how heat energy is distributed within the metal to cause temperature change.

By plugging in the calculated values—heat \(q = 452 \text{ J}\), mass \(m = 36.8 \text{ g}\), and temperature change \(\Delta T = 75.3^{\circ} C\)—we find the specific heat capacity \(c \approx 0.163 \text{ J/g}^{\circ} C\). This tells us for every gram of metal, approximately 0.163 Joules are needed to raise the temperature by \(1^{\circ} C\). Such calculations in calorimetry allow scientists and engineers to design efficiently, considering how materials will behave thermally.