How much heat must be transferred to each kilogram of aluminum to bring it to its melting point, 660°C, from room temperature, 20°C?
To calculate the heat required to bring aluminum to its melting point, we need to use the formula for calculating heat transfer:
Q = m * c * ΔT
Where:
Q is the heat transferred,
m is the mass of the substance (in this case, aluminum),
c is the specific heat capacity of aluminum, and
ΔT is the change in temperature.
Let's plug in the values we have:
Mass of aluminum (m) = 1 kilogram
Specific heat capacity of aluminum (c) = 897 J/(kg°C)
Change in temperature (ΔT) = Final temperature - Initial temperature
Final temperature = 660°C
Initial temperature = 20°C
ΔT = 660°C - 20°C = 640°C
Now, we can calculate the heat transfer:
Q = 1 kg * 897 J/(kg°C) * 640°C
Q ≈ 574,080 J
Therefore, approximately 574,080 joules of heat must be transferred to each kilogram of aluminum to bring it from room temperature to its melting point.