Question # A beaker contains ( 200 mathrm{gm} ) of water. The heat capacity of the beaker is equal to that of ( 20 mathrm{gm} ) of water. The initial temperature of water in the beaker is ( 20^{circ} mathrm{C} ). If ( 440 mathrm{gm} ) of hot water at ( 92^{circ} mathrm{C} ) is poured in it, the final temperature, neglecting radiation loss, will be nearest to

# A beaker contains ( 200 mathrm{gm} ) of water. The heat capacity of the beaker is equal to that of ( 20 mathrm{gm} ) of water. The initial temperature of water in the beaker is ( 20^{circ} mathrm{C} ). If ( 440 mathrm{gm} ) of hot water at ( 92^{circ} mathrm{C} ) is poured in it, the final temperature, neglecting radiation loss, will be nearest to

(1) ( 58^{circ} mathrm{C} )

(2) ( 68^{circ} mathrm{C} )

(3) ( 73^{circ} mathrm{C} )

(4) ( 78^{circ} mathrm{C} )

Solution

Heat lost by hot water = Heat gained by cold water in beaker + Heat absorbed

by

[

begin{array}{c}

440(92-theta)=200 times(theta-20)+20 times(theta-20)

p

end{array}

]

( T=68 circ C )