The figure given below shows the cooling curve of pure wax material after heating. It cools from A to B and solidifies along B D. If L and C are respective values of latent heat and the specific heat of the liquid wax, the ratio L / C is

Question

(A) 40 (B) 80 (C) 100 (D) 20

Tardigrade Admin · Accepted Answer

Let the quantity of heat supplied per minute be Q. Then quantity of heat supplied in 2 min =m C(90-80) In 4 min, heat supplied =2 m C(90-80) ∴ 2 m C(90-80)=m L ⇒ (L/C)=20

Q. The figure given below shows the cooling curve of pure wax material after heating. It cools from $A$ to $B$ and solidifies along $B D$ . If $L$ and $C$ are respective values of latent heat and the specific heat of the liquid wax, the ratio $L / C$ is

40

80

100

20

Let the quantity of heat supplied per minute be $Q$ .
Then quantity of heat supplied in $2 min = m C (90 - 80)$
In $4 min$ , heat supplied $= 2 m C (90 - 80)$
$∴ 2 m C (90 - 80) = m L$
$\Rightarrow \frac{L}{C} = 20$

Q. The figure given below shows the cooling curve of pure wax material after heating. It cools from A to B and solidifies along BD. If L and C are respective values of latent heat and the specific heat of the liquid wax, the ratio L/C is

40

80

100

20

Let the quantity of heat supplied per minute be Q. Then quantity of heat supplied in 2min=mC(90−80) In 4min, heat supplied =2mC(90−80) ∴2mC(90−80)=mL ⇒CL​=20

Q. The figure given below shows the cooling curve of pure wax material after heating. It cools from $A$ to $B$ and solidifies along $B D$ . If $L$ and $C$ are respective values of latent heat and the specific heat of the liquid wax, the ratio $L / C$ is

Let the quantity of heat supplied per minute be $Q$ .
Then quantity of heat supplied in $2 min = m C (90 - 80)$
In $4 min$ , heat supplied $= 2 m C (90 - 80)$
$∴ 2 m C (90 - 80) = m L$
$\Rightarrow \frac{L}{C} = 20$