A 28.4 g sample of an unknown metal is heated to 39.4 °C, then is placed in a calorimeter containing 50.0 g of water. Temperature of water increases from 21.00 °C to 23.00 °C. What is the specific heat of this metal in this problem?
Answers
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
\Rightarrow s_1 =\frac{50\times 2}{28.4\times 16.4}⇒s
1
=
28.4×16.4
50×2
\Rightarrow s_1= 0.214⇒s
1
=0.214 calorie /gram °C
The specific heat of aluminum is 0.214 calorie /gram °C