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What is the heat (energy) required to increase the temperature of a 7.10-kg block of aluminum from 13.0°C to 47.0°C? The specific heat of aluminum is 900 J/kg°C.

User David Wu
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1 Answer

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Final answer:

The energy required to increase the temperature of a 7.10-kg block of aluminum from 13.0°C to 47.0°C is 217,980 joules. The calculation uses the specific heat of aluminum, the mass of the block, and the change in temperature.

Step-by-step explanation:

The student has asked about the heat (energy) required to increase the temperature of a 7.10-kg block of aluminum from 13.0°C to 47.0°C, given that the specific heat of aluminum is 900 J/kg°C. To calculate this, the formula to use is Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat, and ΔT is the change in temperature.

In this case, the change in temperature (ΔT) is 47.0°C - 13.0°C = 34.0°C. Hence, the heat energy (Q) required can be calculated as follows:

Q = (7.10 kg)(900 J/kg°C)(34.0°C)

Q = 217,980 J

Therefore, 217,980 joules of energy is required to increase the temperature of a 7.10-kg block of aluminum from 13.0°C to 47.0°C.

User AriX
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