Answer:
10 °C
Explanation:
From the question given above, the following data were obtained:
Egde length (L) of aluminum = 20 cm
Density of Aluminum = 2.7 g/cm³
Specific heat capacity (C) of aluminum = 0.217 cal/ g°С
Heat (Q) energy = 47000 cal
Change in Temperature (ΔT) =?
Next, we shall determine the volume of the aluminum. This can be obtained as follow:
Egde length (L) of aluminum = 20 cm
Volume (V) of aluminum =?
V = L³
V = 20³
V = 8000 cm³
Thus, the volume of the aluminum is 8000 cm³
Next, we shall determine the mass of the aluminum. This can be obtained as follow:
Density of Aluminum = 2.7 g/cm³
Volume of Aluminum = 8000 cm³
Mass of aluminum =.?
Density = mass/volume
2.7 = mass /8000
Cross multiply
Mass of aluminum = 2.7 × 8000
Mass of Aluminum = 21600 g
Finally, we shall determine the change in temperature of the aluminum as follow:
Specific heat capacity (C) of aluminum = 0.217 Cal/g°С
Heat (Q) energy = 47000 Cal
Mass (M) of Aluminum = 21600 g
Change in Temperature (ΔT) =?
Q = MCΔT
47000 = 21600 × 0.217 × ΔT
47000 = 4687.2 × ΔT
Divide both side by 4687.2
ΔT = 47000 / 4687.2
ΔT = 10 °C
Therefore, the increase in the temperature of the aluminum is 10 °C.
