Thermal Energy Physics Example 1

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Example 1

easy
How much thermal energy is needed to heat 2 kg2 \text{ kg} of water from 20°C20°\text{C} to 80°C80°\text{C}? The specific heat capacity of water is c=4186 J/(kg\cdotp°C)c = 4186 \text{ J/(kg·°C)}.

Solution

  1. 1
    Use the thermal energy formula: Q=mcΔTQ = mc\Delta T.
  2. 2
    Temperature change: ΔT=80−20=60°C\Delta T = 80 - 20 = 60°\text{C}.
  3. 3
    Q=2×4186×60=502,320 J≈502 kJQ = 2 \times 4186 \times 60 = 502{,}320 \text{ J} \approx 502 \text{ kJ}

Answer

Q≈502 kJQ \approx 502 \text{ kJ}
Thermal energy (heat) required to change an object's temperature depends on its mass, specific heat capacity, and the temperature change. Water has a very high specific heat, requiring lots of energy to heat.

About Thermal Energy

The total kinetic energy of all particles (atoms and molecules) in an object due to their random motion.

Learn more about Thermal Energy →

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