Thermal Energy Physics Example 3

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Example 3

medium
A 3 kg3 \text{ kg} aluminum block (c=900 J/(kg\cdotp°C)c = 900 \text{ J/(kg·°C)}) absorbs 54,000 J54{,}000 \text{ J} of thermal energy. What is the temperature increase?

Solution

  1. 1
    Rearrange Q=mcΔTQ = mc\Delta T for ΔT\Delta T.
  2. 2
    ΔT=Qmc=54,0003×900=54,0002700=20°C\Delta T = \frac{Q}{mc} = \frac{54{,}000}{3 \times 900} = \frac{54{,}000}{2700} = 20°\text{C}

Answer

ΔT=20°C\Delta T = 20°\text{C}
The temperature rise is inversely proportional to both mass and specific heat capacity. The same energy would heat a smaller mass of aluminum by a larger amount.

About Thermal Energy

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

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