Question

Substances \(\mathbf{A}\) and \(\mathbf{B}\) are the same mass and at the same initial temperature. When they are heated, the final temperature of A is \(55^{\circ} \mathrm{C}\) higher than the temperature of \(\mathrm{B}\). What does this tell you about the specific heats of \(\mathbf{A}\) and \(\mathbf{B}\) ?

Short answer

The specific heat of substance \( A \) is higher than that of substance \( B \).

Step-by-step solution

Understanding Specific Heat

Specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius. Mathematically, it is given by the formula: \[ q = m \times c \times \triangle T \] where \( q \) is the heat absorbed or released, \( m \) is the mass, \( c \) is the specific heat and \( \triangle T \) is the change in temperature.

Express the Given Information

Given that substances \( \textbf{A} \) and \( \textbf{B} \) are of the same mass and initial temperature, and that the final temperature of \( \textbf{A} \) is \( 55^{\text{°C}} \) higher than that of \( \textbf{B} \), we denote the temperatures as: \[ T_f(A) = T_i + \triangle T(A) \] \[ T_f(B) = T_i + \triangle T(B) \] Hence, \[ \triangle T(A) = \triangle T(B) + 55 \]

Relate Heat Absorbed to Specific Heat

For each substance, the heat absorbed can be expressed as: \[ q(A) = m \times c_A \times \triangle T(A) \] \[ q(B) = m \times c_B \times \triangle T(B) \] Given the same heating conditions, \( q(A) = q(B) \).

Substitute and Solve

Since the heat absorbed by both substances is the same: \[ m \times c_A \times \triangle T(A) = m \times c_B \times \triangle T(B) \] The masses cancel out: \[ c_A \times \triangle T(A) = c_B \times \triangle T(B) \] Substitute \( \triangle T(A) = \triangle T(B) + 55 \): \[ c_A \times (\triangle T(B) + 55) = c_B \times \triangle T(B) \]

Compare Specific Heats

Isolate \( c_A \): \[ c_A \times \triangle T(B) + 55 c_A = c_B \times \triangle T(B) \] Divide by \( \triangle T(B) \) (assuming \( \triangle T(B) \) is not zero): \[ c_A + \frac{55 c_A}{\triangle T(B)} = c_B \] This equation shows that \( c_A > c_B \) because there is an additional term, indicating that substance \( A \) has a higher specific heat than substance \( B \).

Key concepts

specific heat

Specific heat is a key concept in understanding how different substances respond to heat. It refers to the amount of heat required to raise the temperature of one unit mass of a substance by one degree Celsius. The formula for calculating the heat absorbed or released by a substance is: \[ q = m \times c \times \triangle T \] where

• \( q \) is the heat absorbed or released
• \( m \) is the mass of the substance
• \( c \) is the specific heat
• \( \triangle T \) is the change in temperature

Each substance has a unique specific heat, which affects how quickly it heats up or cools down. In the exercise, this concept helps us compare how much heat substances A and B can absorb to change their temperatures.

heat transfer

Heat transfer is the movement of thermal energy from one object to another. Heat can be transferred in three ways: conduction, convection, and radiation. In our exercise, the heat is transferred to substances A and B, causing their temperatures to rise. Understanding heat transfer helps us comprehend why different materials heat up at different rates:

• Conduction involves direct contact, where heat transfers through materials touching each other.
• Convection occurs in fluids, like air or water, where heated particles move and carry energy.
• Radiation involves heat transfer through electromagnetic waves, even in a vacuum.

In the problem, since both substances A and B are heated under the same conditions, they absorb the same amount of heat. This allows us to focus on the specific heat to explain the temperature differences.

temperature change

Temperature change, represented as \( \triangle T \), is how much the temperature of a substance increases or decreases. The relationship between heat, specific heat, and temperature change can be seen in the formula: \[ q = m \times c \times \triangle T \] In the problem, it’s given that the final temperature of substance A is 55°C higher than that of substance B. This difference helps us compare their specific heats. Since specific heat determines the amount of heat needed to change a substance's temperature by 1°C, the formula \[ c_A \times (\triangle T_B + 55) = c_B \times \triangle T_B \] leads us to conclude that substance A must have a higher specific heat than substance B. This is evident because substance A requires more heat to achieve a higher temperature increase than substance B.