Question

A manufacturer of helicopter blades heat-treats a part as one of the steps of the processing procedure. Heat-treating involves heating the metal up to a

highly specific temperature and then quenching, or cooling, it down quickly. The blades are made from an aluminum alloy and have a mass of 273 kg.
Water is normally used for the quenching phase. If the water is at a temperature of 22°C, how much water (in kg) would be needed to cool the blade
from 300°C to at least 50°C?
External Data
The Cp of water is 4,180 J/kg °C, and the C, of aluminum alloy is 900 J/kg °C. The equation for calculating heat gained or lost by a material is q = mCAT.

Answer 1

To determine the mass of water needed to cool the helicopter blades from 300°C to 50°C, the heat lost by the aluminum blades (calculated using mass, specific heat, and temperature change) must equal the heat gained by the water. This allows calculating the required mass of water using the specific heat and temperature change of water.

Step-by-step explanation:

To find the mass of water needed to cool the helicopter blades made from an aluminum alloy from 300°C to 50°C, we use the principle of conservation of energy. The heat lost by the blades must equal the heat gained by the water, meaning:

Heat loss by aluminum = Heat gain by water

Qloss = mblade × caluminum × ΔTblade
Qgain = mwater × cwater × ΔTwater

Given:

• mblade = 273 kg
• caluminum = 900 J/kg°C
• ΔTblade = (50°C - 300°C) = -250°C
• cwater = 4180 J/kg°C
• ΔTwater = (50°C - 22°C) = 28°C

Calculating the heat loss by the aluminum blades:

Qloss = (273 kg) × (900 J/kg°C) × (-250°C)

Then, calculating the required mass of water having calculated Qloss:

mwater = Qloss / (cwater × ΔTwater)