## Chapter 1 Physical Environment Solved Numerical Problems

**Problem 1 How much heat will be required to raise the temperature of 200 grams of water from 30°C to 90°C? (Given: specific heat of water is 1 cal/g.°C)**

**Solution:**

**Given:**

Temperature of 200 grams of water from 30°C to 90°C

Initial temperature of the water, t = 30°C

The final temperature of the water, t2 = 90°C

Change in temperature,

(t2-t1) = (90-30)°C = 60°C

Mass of water, m = 200 g

Specific heat of water, s = 1 cal/g.°C

The heat required = m.s. (t2 – t1) = 200 x 1 x 60 cal = 12000 cal.

**Problem 2. A brass utensil of 400 g weight at 100°C cools down to 25°C by radiating heat. Calculate the amount of heat released by the brass utensil. (Given: specific heat of brass = 0.09 cal/g.°C)**

**Solution :**

**Given:**

A brass utensil of 400 g weight at 100°C cools down to 25°C by radiating heat.

The initial temperature of brass, t1 = 100°C

The final temperature of brass, t2 = 25°C

Change in temperature, (t1-t2) = ( 100-25)°C = 75°C

Mass of brass, m = 400 g Specific heat of brass,

s = 0.09 cal/g.°C Heat rejected by the brass utensil = m.s.(t1 -t2)

= 400 x 0.09 x 75 cal = 2700 cal

**Read And Learn More WBBSE Solutions For Class 8 School Science Solved Numerical Problems**

**Problem 3. 100 grams of water is taken at 20°C temperature. If 50 grams of a substance at 100°C is dropped in that water, the final temperature becomes 50°C. Calculate the specific heat of that substance.**

**Solution:**

**Given:**

100 grams of water is taken at 20°C temperature. If 50 grams of a substance at 100°C is dropped in that water, the final temperature becomes 50°C.

Heat gained by water,

= 100 x l x (50 – 20) cal = 3000 cal Heat released by the substance,

H2 = 50 x s x (100 – 50) cal,

where “s” is the specific heat of the substance

We know that, heat rejected = heat gained or, H2 = H1

or, 50 x s x (100 – 50) = 3000 cal

or,s =3000/50×50 =1.2 cal/g°.C

## Chapter 1 Physical Environment Solved Numerical Problems

**Problem 1. The latent heat for melting ice at 0°C and 1 atm is 80 cal/g. How much heat has to be supplied to completely convert 200g ice at 0°C to water at 0°C**

**Solution:**

**Given:**

Latent heat for melting of ice at 0°C and 1 atm = 80 cal/g

The heat required to completely convert 1 g ice at 0°C to water at 0°C = 80 cal

So, the heat is required to completely convert 200 g ice at 0°C to water at 0°C = (80 x 200) cal = 16000 cal.

**Problem 2. The latent heat for melting ice at 0°C and 1 atm is 80 cal/g. Calculate its value in the SI units.
**

**Solution:**

**Given:**

Latent heat for melting of ice at 0°C and 1 atm is 80 cal/g

The heat required to completely convert 1 g ice at 0°C to water at 0°C = 80 cal

So, the heat required to completely convert

1000 g ice at 0°C to water at 0°C

= (80 x 1000) cal = 80000 cal

We know that 1 cal = 4.2 J

Hence, 80000 cal

= (80000 x 4.2) J = 336000 J So, latent heat of melting of ice = 336000 J/kg