7. A 28.4 g sample of aluminum is heated to 39.4 °C, then is placed in a calorimeter
1 answer:
Answer:
The specific heat of aluminum is 0.214 calorie /gram °C
Explanation:
Specific heat:The heat requires to increase 1°C temperature of that object per unit gram.
When the aluminum comes in contact of of water then
The heat lose by aluminum = The heat gain by the water.
Q₁=Q₂
Where,
Q = m ×s× ΔT
m= mass of an object
s= specific heat of the object
ΔT= change of temperature.
The mass of aluminum(m₁)= 28.4 gram,
ΔT for aluminum=(initial temperature - final temperature)= (39.4-23)°C=16.4°C
s₁=?
The mass of water(m₂) = 50 gram
ΔT for water =( final temperature-initial temperature)=(23-21)°C= 2°C
S₂=1 calorie /gram °C
Q₁=Q₂
⇒m₁s₁ΔT=m₂s₂ΔT
⇒(28.4×s₁×16.4)= 50×1×2
calorie /gram °C
The specific heat of aluminum is 0.214 calorie /gram °C
You might be interested in
Number of coulombs of positive charge in 250cm^3 water is 1.3×10^7 C
The volume of 250 cm^3 corresponds to a mass of 250 g since the density of water is 1.0 g/cm^3
This mass corresponds to 250/18 = 14 moles since the molar mass of water is 18. There are ten proton (each with charge q = +e) in each molecule of So,
Q = 14NA q =14(6.02×10^23)(10)(1.60×10^−19C) = 1.3×10^7 C.
Mass is the quantity of matter in a physical body. It is also a measure of the body's inertia, the resistance to acceleration when a net force is applied. An object's mass also determines the strength of its gravitational attraction to other bodies.
Learn more about mass here:
#SPJ4