Answer:
1. The specific heat of the meatal is approximately 0.266 J/(g·°C)
2. 46.6334 kJ
Explanation:
1. The given mass of the sample of metal, m = 110.0 g
The temperature change of the metal when heat is added, ΔT = 10.0°C
The amount of head that was added, ΔH = 70.0 Calories = 292.88 Joules
The formula for heat heat capacity of a material is given as follows
Q = m × c × ΔT
Where;
Q = The quantity of heat added = 292.88 Joules
m = The mass of the material = 110.0 g
ΔT = The temperature change = 10.0°C
c = The specific heat capacity of the material
We get;
c = ΔH/(m × ΔT)
Which gives;
c = 292.88 J/(110.0 g × 10.0°C) ≈ 0.266 J/(g·°C)
2. The given mass of aluminum, m = 550.0 g
The initial temperature of the aluminum, T₁ = 14.0°C
The final temperature of the aluminum, T₂ = 108.0°C
The specific heat of aluminum, c = 0.902 J/(g·°C)
We get;
Q = m × c × (T₂ - T₁)
Q = 550.0 g × 0.902 J/(g·°C) × (108.0°C - 14.0°C) = 46,633.4 joules = 46.6334 kJ
The heat that must be supplied to 550.0 g of aluminum pan to raise the temperature of the pan from 14.0°C to 108.0°C, Q = 46.6334 kJ.
