Question

The amount of heat energy required to vapourise 2 grams of ice at 0°C is? (latent heat of fusion of ice = 80
cal/g and latent heat of vaporization of water = 540 cal/g)
1240 cal
1340 cal
800 cal
200 cal​

Answer 1

Answer:

Given:

⇒ Mass (m) = 2 g

⇒ Temperature ($\sf \Delta T$) = 100°C (100 - 0)

⇒ Latent heat of fusion of ice = 80 cal/g

⇒ Latent heat of vaporization of water = 540 cal/g

To Find:

⇒ The amount of heat energy required to vapourise 2 grams of ice

Solution:

We know that,

Heat required to vapourise ice

$\blue{\boxed{\sf Q = \sf mL_i_c_e + mC \Delta T +mL_w_a_t_e_r}}$

Given that,

m = 2 g

$\sf \Delta T$ = 100°C

Specific heat (C) = 1 cal/g

$\sf L_i_c_e = 80\;cal/g$

$\sf L_w_a_t_e_r = 540\;cal/g$

Substituting the the above values in the formula.

We get,

$\sf \implies Q = (1 \times 80) + 2(1)(100) +(2 \times 540)\;cal$

$\sf \implies Q = 160 + 200 + 1080\;cal$

$\sf \implies Q = 1440\;cal$

Hence,

The amount of heat energy required to vapourise 2 grams of ice is 1440 cal

{this is correct, options are wrong}

Answer 2

Answer:

Explanation:

Given: m = 150g

L = 540 cal/g

To find: Q = ?

Formula: Q = mL

Solution: Q = 150 × 540

= 81000 cal

Heat energy needed is 81000 cal.