$$10$$ gm of ice at $$-20oC$$ is dropped into a calorimeter containing $$10$$ gm of water at $$10oC$$; the specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain

Question

Infinity Learn · Accepted Answer

$$Q1=10$$×$$1$$×$$10=100calQ2=10$$×$$0.5$$[$$0$$−$$($$−$$20)$$]$$+10$$×$$80=(100+800)cal=900calAs$$ $$Q1$$ < $$Q2$$, so ice will not completely melt and final temperature $$=$$ $$0oC$$.As heat given by water in cooling upto $$0oC$$ is only just sufficient to increase the temperature of ice from $$-20oC$$ to $$0oC$$ hence mixture in equilibrium will consist of $$10$$ gm of ice and $$10$$ gm of water, both at $$0oC$$.

10 gm of ice at -20^oC is dropped into a calorimeter containing 10 gm of water at 10^oC; the specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

10 gm of ice at -20oC is dropped into a calorimeter containing 10 gm of water at 10oC; the specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

10 gm of ice at -20^oC is dropped into a calorimeter containing 10 gm of water at 10^oC; the specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain