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:

Question

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: (A)  32° C  (B)  16° C  (C)  8° C  (D)  24° C

Tardigrade Admin · Accepted Answer

Let final temperature be θ Now heat taken by ice =m1L+m1c1θ 1 =5× 80+5× 1(θ -0) =400+5θ ...(1) Heat given by water at 40Â°C =m2l2θ 2=20× 1× (40o-θ ) .. .(2) Heat given = Heat taken 800-20θ =400+5θ Or 25θ =400, Or θ =(400/25) =16° C

Q. 5 g of ice at $0^{\circ} C$ is dropped in a beaker containing 20 g of water at $40^{\circ} C$ . The final temperature will be:

$32^{\circ} C$

$16^{\circ} C$

$8^{\circ} C$

$24^{\circ} C$

Q. 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:

32∘C

16∘C

8∘C

24∘C

Let final temperature be θ Now heat taken by ice =m1​L+m1​c1​θ1​ =5×80+5×1(θ−0) =400+5θ ...(1) Heat given by water at 40°C =m2​l2​θ2​=20×1×(40o−θ) .. .(2) Heat given = Heat taken 800−20θ=400+5θ Or 25θ=400, Or θ=25400​ =16∘C

Q. 5 g of ice at $0^{\circ} C$ is dropped in a beaker containing 20 g of water at $40^{\circ} C$ . The final temperature will be:

$32^{\circ} C$

$16^{\circ} C$

$8^{\circ} C$

$24^{\circ} C$