Specific Heat

The specific heat formula calculates the amount of heat transferred into and out of a system. This equation provides a quantitative relationship between heat transfer, substance’s mass, specific heat, and temperature change and is shown as follows:

q = mcΔT

In this formula, q represents the amount of heat transferred. It is measured in Joules (J) or calories (cal). The variable m denotes the mass of the substance being heated or cooled. It is typically measured in grams (g) or kilograms (kg). The specific heat of the substance is represented by the variable c. It is measured in Joules per gram per degree Celsius (J/g-°C) or Joules per kilogram per degree Celsius (J/kg-°C). Finally, ΔT represents the temperature change experienced by the substance. It can be calculated by subtracting the initial temperature from the final temperature and is usually expressed in degrees Celsius or Kelvin.

This formula assumes no phase change occurs during heating or cooling processes and applies specifically to substances with constant specific heat within a given temperature range.

Instead of unit mass, if one mole of a substance is used, physicists use molar specific heat, represented by C.

The specific heat of water plays a fundamental role in various natural processes and everyday life. It refers to the amount of heat energy required to raise the temperature of a unit mass of water by 1°C. Its value is 4.184 J/g-°C or 1 cal/g-°C.

Water’s high specific heat capacity is one of its most remarkable properties. Water has an exceptionally high capacity to absorb and retain heat energy compared to other substances. It can absorb a significant amount of heat without undergoing large temperature changes.

This property is significant in thermoregulation in nature and human-made systems. In nature, bodies of water such as oceans, lakes, and rivers act as thermal regulators by absorbing excess heat during warmer periods and releasing it slowly during cooler periods. It helps maintain stable temperatures in surrounding areas and supports diverse ecosystems.

In human-made systems, understanding the specific heat of water is crucial for designing efficient heating and cooling systems. Water’s ability to store large amounts of thermal energy makes it an ideal medium for transferring heat between different areas or objects while minimizing temperature fluctuations.

Below is a table listing the specific heats of the mentioned substances:

Substance	Specific Heat (J/g-°C)
Water	4.184
Aluminum	0.897
Copper	0.385
Ice	2.090
Iron	0.449
Air	1.006
Gold	0.129
Silver	0.240
Steam	2.020
Steel	0.466
Brass	0.380
Zinc	0.388
Lead	0.128
Glass	0.840
Mercury	0.140
Tin	0.227
Sand	0.830
Nickel	0.444

Unlike water, metals have lower specific heat. It means these metals require less energy to increase their temperature than an equal mass of water.

While both terms are related to the amount of heat energy required to raise the temperature of a substance, specific heat and heat capacity differ in their units of measurement and how they are applied.

Specific heat refers to the amount of heat energy needed to raise the temperature of a unit mass of a substance by one degree Celsius or Kelvin. It is an intrinsic property that varies from one material to another. The specific heat values allow physicists to compare how different substances absorb and retain heat differently.

On the other hand, heat capacity measures the amount of thermal energy required to raise the temperature of an entire object or system by one degree Celsius or Kelvin. Unlike specific heat, which focuses on a unit mass, heat capacity considers all components and mass within a given system. Hence, it is an extrinsic property. Heat capacity is typically measured in Joules per degree Celsius (J/°C) or Joules per Kelvin (J/K).

Problem 1:  A 200 g piece of aluminum initially at 80°C is dropped into 400 mL of water at 20°C. If the final temperature of the system is 40°C, calculate the heat transferred to the water. The specific heat of aluminum is 0.9 J/g-°C, and the specific heat of water is 4.18 J/g-°C.

Solution:

The heat transfer equation is given by:

For aluminum:

\[ q_{\text{Al}} = m_{\text{Al}} \cdot c_{\text{Al}} \cdot \Delta T_{\text{Al}} \]

\[ q_{\text{Al}} = (200 \, \text{g}) \cdot (0.9 \, \text{J/g°C}) \cdot (40 \, \text{°C} – 80 \, \text{°C}) \]

\[ q_{\text{Al}} = -7200 \, \text{J} \]

For water:

\[ q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}} \]

\[ q_{\text{water}} = (400 \, \text{g}) \cdot (4.18 \, \text{J/g°C}) \cdot (40 \, \text{°C} – 20 \, \text{°C}) \]

\[ q_{\text{water}} = 33440 \, \text{J} \]

The total heat transfer is the sum of the heat transferred to aluminum and water:

\[ q_{\text{total}} = q_{\text{Al}} + q_{\text{water}} \]

\[ q_{\text{total}} = -7200 \, \text{J} + 33440 \, \text{J} \]

\[ q_{\text{total}} = 26240 \, \text{J} \]

Problem 2: A 150 g piece of copper is heated from 20°C to 100°C. Calculate the heat energy absorbed by the copper. The specific heat of copper is 0.39 J/g-°C.

Solution:

The heat transfer equation is given by:

\[ q = m \cdot c \cdot \Delta T \]

\[ q_{\text{copper}} = m_{\text{copper}} \cdot c_{\text{copper}} \cdot \Delta T_{\text{copper}} \]

\[ q_{\text{copper}} = (150 \, \text{g}) \cdot (0.39 \, \text{J/g°C}) \cdot (100 \, \text{°C} – 20 \, \text{°C}) \]

\[ q_{\text{copper}} = 3510 \, \text{J} \]