Consider a 10 gram sample of a liquid with specific heat 2 j/gk. by the addition of 400 j, the liquid increases its temperature by 10 k and then boils. half of the liquid boils away before all the heat is used up. what is the heat of vaporization of the liquid? a.10 j/gb.20 j/gc.80 j/gd.200 j/ge.40 j/g© 2018.grand canyon university. all rights reserved.

40 J/g is the heat of vaporization of the liquid.

Option D

Explanation:

Given that mass of liquid sample: m = 10 g

And, Specific heat of the liquid: S = 2 J/g K

Also, the increase in the temperature of the liquid,  

Therefore, the total amount of heat energy required is given by:

According to the given data in the question,

Total heat energy supplied, q = 400 J

Rest of heat would be

Now, 200 J vaporizes the mass, half of the liquid from full portion boiled away. So,

Latent heat of vaporization of the liquid is . It can be calculated as below,

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards