An 8000-kg aluminum flagpole 100-m long is heated by the Sun from a temperature of 10°C to 20°C. Find the heat transferred (in J) to the aluminum if the specific heat of aluminum is 0.215 cal/g⋅°C.

Question

Quizplus · Accepted Answer

The heat transferred to the aluminum flagpole can be calculated using the formula:

Q = mcΔT

where Q is the heat transferred, m is the mass of the aluminum, c is the specific heat of aluminum, and ΔT is the change in temperature.

Here, m = 8000 kg, c = 0.215 cal/g⋅°C = 0.215 J/g⋅°C, and ΔT = 20°C - 10°C = 10°C = 10 K.

We need to convert the mass from kg to g and the specific heat from J/g⋅°C to J/Kg⋅°C:

m = 8000 kg = 8.0 × 10^6 g
c = 0.215 J/g⋅°C × (1 kg/1000 g) = 0.000215 J/Kg⋅°C

Substituting these values into the formula, we get:

Q = (8.0 × 10^6 g) × (0.000215 J/Kg⋅°C) × 10 K
  = 1.72 × 10^3 J

Therefore, the heat transferred to the aluminum flagpole is 1.72 × 10^3 J, which is equivalent to 7.2 × 10^17 cal.

The closest answer choice to this value is B, 7.2 × 10^17 cal.

An 8000-Kg Aluminum Flagpole 100-M Long Is Heated by the Sun