Question

1.how much steam must be passed into a mixture of Ice and water in order to melt 10 gram of ice.(Quantity of heat) ​

Answer 1

Answer: 1.25 gm of steam is required.

Given: Mass of ice = 10 gm

To Find: Quantity of steam required.

Explanation:

Step 1: Q = mL

where, Q= heat energy

M = per mass released

L = Latent heat of vaporization

Step 2: The final temperature of steam must match that of ice, which is zero, in order to achieve thermal equilibrium.

The heat released by steam as it transitions from gas to liquid is given by Q=mL, where L is the latent heat of vaporization and m is the mass of the matter.

During a constant-temperature process – typically a first-order phase change – latent heat is emitted or consumed by a body or a thermodynamic device.

Water has a latent heat of evaporation of 540 cal/g and a latent heat of freezing of 80 cal/g, for example.

Step 3: Let x gram of steam is required. So, latent heat of vaporization of steam is

$Q_1=mL=540x$

The heat emitted by boiling water changes its temperature from 100 °C to 0°C is given as

$Q_2=mC\triangle T=100x$

Total heat emitted by x gram of steam to change itself at 0°C of water is

$Q=Q_1+Q_2=540x+100x=640x$

As we know that latent heat of fusion of 10 gram of ice is

$Q=mL_f\\\\640 x= 10\times80=800\\\\x=\frac{800}{640} =1.25\ gm$

So, we can say that 1.25 gram of steam is required to melt 10 gram of ice.

For more questions follow the link given below

https://brainly.in/question/39679177

https://brainly.in/question/40007651

#SPJ1