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define-temperature-and-specific-heat

🚀 Temperature is a measure of the average kinetic energy of the particles in a substance. It indicates how hot or cold an object is. Specific heat is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). The specific heat capacity varies from one material to another and is crucial in understanding how different substances respond to heat.

Theory Explanation

Understanding Temperature

Temperature is a scalar quantity that represents the thermal state of a system. It is measured in degrees Celsius (°C), Kelvin (K), or Fahrenheit (°F). The Kelvin scale is the SI unit of temperature, where 0 K is absolute zero, the point at which all molecular motion ceases.

\[ T(K) = t(°C) + 273.15 \]
Defining Specific Heat

Specific heat (c) is defined as the amount of heat (Q) required to change the temperature (ΔT) of a unit mass (m) of a substance. The formula is given by: Q = mcΔT, where Q is the heat added, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

\[ Q = mcΔT \]
Calculating Heat Transfer

When heat is added to or removed from a substance, its temperature changes. The relationship between heat transfer, mass, specific heat, and temperature change can be used to calculate the heat involved in physical processes such as heating, cooling, and phase changes.

\[ Q = mc(T_f - T_i) \]

Key Points

  • 🎯 Temperature measures the average kinetic energy of particles.
  • 🎯 Specific heat is unique to each substance and affects how it absorbs heat.
  • 🎯 The formula Q = mcΔT is fundamental in calculating heat transfer.

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Examples:💡

Calculate the heat required to raise the temperature of 2 kg of water from 20°C to 100°C. (Specific heat of water = 4.18 J/g°C)

Solution:

Step 1: Identify the mass (m) of water: 2 kg = 2000 g.

\[ m = 2000 g \]

Step 2: Determine the specific heat (c) of water: c = 4.18 J/g°C.

\[ c = 4.18 J/g°C \]

Step 3: Calculate the change in temperature (ΔT): ΔT = T_f - T_i = 100°C - 20°C = 80°C.

\[ ΔT = 80°C \]

Step 4: Use the formula Q = mcΔT to find the heat (Q): Q = 2000 g * 4.18 J/g°C * 80°C = 668800 J.

\[ Q = 2000 g * 4.18 J/g°C * 80°C = 668800 J \]

How much heat is released when 1 kg of ice at 0°C melts to water at 0°C? (Latent heat of fusion of ice = 334 J/g)

Solution:

Step 1: Identify the mass (m) of ice: m = 1 kg = 1000 g.

\[ m = 1000 g \]

Step 2: Determine the latent heat of fusion (L) of ice: L = 334 J/g.

\[ L = 334 J/g \]

Step 3: Calculate the heat (Q) released during melting: Q = mL = 1000 g * 334 J/g = 334000 J.

\[ Q = 1000 g * 334 J/g = 334000 J \]

Common Mistakes

  • Mistake: Confusing temperature with heat; temperature is not a measure of heat but of kinetic energy.

    Correction: Remember that temperature indicates how hot or cold something is, while heat refers to energy transfer.

  • Mistake: Using incorrect units for specific heat or mass in calculations.

    Correction: Always ensure that mass is in grams when using specific heat in J/g°C, or convert to kg if using J/kg°C.

  • Mistake: Neglecting to account for the direction of heat transfer (adding vs. removing heat).

    Correction: Clearly define whether heat is being added or removed in problems to avoid sign errors in calculations.

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