High School

The temperature of a 5.00 kg lead brick is increased by 525 °C. If the specific heat capacity of lead is 128 J/kg°C, calculate the heat energy absorbed by the lead brick.

Answer :

To calculate the heat energy absorbed by the lead brick, we will use the formula for heat energy (Q) given by the equation:

[tex]Q = m \cdot c \cdot \Delta T[/tex]

Where:

  • [tex]Q[/tex] is the heat energy absorbed (in joules)
  • [tex]m[/tex] is the mass of the substance (in kilograms)
  • [tex]c[/tex] is the specific heat capacity of the substance (in J/kg°C)
  • [tex]\Delta T[/tex] is the change in temperature (in °C)

Given Data:

  • Mass of lead brick, [tex]m = 5.00 \text{ kg}[/tex]
  • Specific heat capacity of lead, [tex]c = 128 \text{ J/kg°C}[/tex]
  • Change in temperature, [tex]\Delta T = 525°C[/tex]

Step-by-Step Calculation:

  1. Substitute the known values into the heat energy formula:

    [tex]Q = 5.00 \text{ kg} \times 128 \text{ J/kg°C} \times 525°C[/tex]

  2. Calculate the product:

    [tex]Q = 5.00 \times 128 \times 525[/tex]

  3. Perform the multiplication:

    [tex]Q = 336000 \text{ J}[/tex]

Therefore, the heat energy absorbed by the lead brick is [tex]336,000 \text{ joules}[/tex].

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