(455) Specific heat capacity [IB Physics SL/HL]

Specific Heat Capacity

• Definition: Energy needed to raise the temperature of 1 kg of a substance by 1° Kelvin.

• Equation: ( Q = m \cdot c \cdot \Delta T )

  • ( Q ): heat (Joules)

  • ( m ): mass (kilograms)

  • ( \Delta T ): change in temperature (°C or K)

  • ( c ): specific heat capacity (not the speed of light)

Key Concepts

• Heat and Energy: Heat is a form of energy; increased heat leads to increased temperature.

• Materials Variation: Different materials have different specific heat capacities; it's easier to change the temperature of substances with low specific heat capacities.

Rearranging the Equation

• To solve for ( c ): ( C = \frac{Q}{m \cdot \Delta T} )

• Units for Specific Heat Capacity: ( C ) is measured in J/(kg·°C).

Heat Exchange Principle

• Heat Gained vs Heat Lost: ( Q_{\text{gained}} = Q_{\text{lost}} )

  • Use this principle to analyze systems where energy is transferred between objects.

Example Problem: Tea and Cup

• Given: 200 g of tea at 95° C into a 150 g glass cup at 25° C.

• Specific Heat Capacities: Tea (like water): 4186 J/(kg·°C); Glass: 840 J/(kg·°C).

Setting Up the Equation

• Define the heat exchange:

  • ( Q_{T} = m_{T} \cdot c_{T} \cdot \Delta T_{T} )

  • ( Q_{cup} = m_{cup} \cdot c_{cup} \cdot \Delta T_{cup} )

• Mass Conversions: 200 g = 0.2 kg; 150 g = 0.15 kg.

Temperature Changes

• Since equilibrium temperature ( T ) is sought:

  • For tea, ( \Delta T_T = 95 - T )

  • For cup, ( \Delta T_{cup} = T - 25 )

Solving the Equation

• Substitute the values into ( Q_{lost} = Q_{gained} ):

  • ( (0.2 \cdot 4186 \cdot (95 - T)) = (0.15 \cdot 840 \cdot (T - 25)) )

• Solve for ( T ) step by step using algebra.

• Final answer after calculations: Equilibrium Temperature = 86° C (to 2 significant figures).

Tips for Problem Solving

• Always track which object is losing and which is gaining energy.

• Set up equations systematically based on energy transfer principles.