Question

A spare of 100gram mass is heated to raise it's temperature of its temperature to 100°C and realesd in water of mass 195gram and temperature 20°C in a copper if a mass of calorie meter is 30°C... What will be the maximum temperature..?​

Answer 1

Answer:

Given : Specific heat of copper = 0.1 cal/g ∘C

Specific heat of calorimeter = 0.1 cal/g∘C

Mass of copper sphere = 100 g

Mass of water = 195 g

Mass of calorimeter = 50 g

Solution: Suppose the copper ball, water and the calorimeter attain final temperature T. Heat lost by solid object = heat gained by water in calorimeter + heat gained by calorimeter

Here, heat lost by the copper ball = mass of the copper x specific heat of copper x decrease in temperature of ball

Q=100 x 0.1 x (100-T)

Similarly, Heat gained by the water = mass of the water x specific heat of water x increase in its temperature

Q1=195×1×x(T−20)

Heat gained by the calorimeter = mass of the calorimeter x specific heat x increase in its temperature

Q2 = 50 x 0.1 x (T-20)

Now, Q=Q1+Q2

00 x 0.1 x (100-T) = 195 x 1 x (T-20 ) + 50 x 0.1 x (T-20)

10(100-T)= 195 (T-20) + 5(T-20)

10 (100 - T ) = 200 (T-20)

1000-10T = 200T-4000

210 T = 5000

T=23.8∘C

The maximum temperature of water will be 23.8∘C