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Answer :
Final answer:
The energy required to raise the temperature of a 3 kg piece of lead from 40°C to 72°C is 12.288 kilojoules.
Explanation:
The student is asking how much energy is required to raise the temperature of a 3 kg piece of lead from 40°C to 72°C. To calculate this, we can use the formula:
Q = mcΔT,
where Q is the energy in joules, m is the mass in kilograms, c is the specific heat capacity, and ΔT is the change in temperature in degrees Celsius.
Given:
- m = 3 kg
- c = 128 J/kg°C
- ΔT = 72°C - 40°C = 32°C
Plugging in these values:
Q = 3 kg × 128 J/kg°C × 32°C
Q = 12288 J
Since 1 kilojoule (kJ) is equal to 1000 joules (J), we divide the result by 1000 to convert it:
Q = 12.288 kJ
Therefore, the energy required to raise the temperature of the 3 kg piece of lead by 32°C is 12.288 kJ.
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