`0^(@)C` তে 5 গ্রাম বরফ একটি বীকারে ফেলে দেওয়া হয় যাতে 20 গ্রাম জল থাকে `40^(@)C`।

5 g of ice at 0^(@)C is dropped in a beaker containing 20 g of water at 40^(@)C . The final temperature will be

5.0 L water dissolves 80mL N_(2) gas at 10 bar and 25^(@)C . The volume of water needed to dissolve 40mL N_(2) gas at 20 bar at 25^(@)C , is:

A body cools from 50^(@)C to 40^(@)C is 5 minutes when its surrounding are at a constant temperature of 20^(@)C . How long will it take for it to cool by another 10^(@)C ?

A refrigerator freezes 5 kg of water at 0^(@)C into ice at 0^(@)C in a time internal of 20 minutes. Assume that the room temperature is 20^(@)C . Calculate the minimum power needed to accomplish it.

A body cools from 50.0^(@)C to 48^(@)C in 5s. How long will it take to cool from 40.0^(@)C to 39^(@)C ? Assume the temperature of surroundings to be 30.0^(@)C and Newton's law of cooling to be valid.

A body cools from 50.0^(@)C to 49.9^(@)C in 5s. How long will it take to cool from 40.0^(@)C to 39.9^(@)C ? Assume the temperature of the surroundings to be 30.0^(@)C and Newton's law of cooling to be valid:

A body cools at the rate of 0.6^(@)C//s when it is 40^(@)C above the surrounding. Then the rate of cooling when it is at 20^(@)C above the same surrounding is

A pendulum clock consists of an iron rod connected to a small, heavy bob. If it is designed to keep correct time at 20^(0) C , how fast or slow will it go in 24 hours at 40^(0) C? Coefficient of linear expansion of iron = 1.2 xx 10^(-5) ^0 C^(-6) .

When a body is placed in surroundings at a constant temperature of 20^(@)C and heated by a 10W heater, its temperature remains constant at 40^(@)C . If the temperature of the body is now raised from 20^(@)C to 80^(@)C in 5 min at a uniform rate, the total heat it will lose to the surroundings will be