Question

7. A 28.4 g sample of aluminum is heated to 39.4 °C, then is placed in a calorimeter

Answer 1

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

$\Rightarrow s_1 =\frac{50\times 2}{28.4\times 16.4}$

$\Rightarrow s_1= 0.214$ calorie /gram °C

The specific heat of aluminum is 0.214 calorie /gram °C