Take Cice - 0-53 cal-g - C- Cwater - 1-0 cal-g - C- Lfwater - 80 cal-g and Lvwater - 529 cal-g- Equal masses of ice at 0- C and water are in contact- Find the temperature of water needed to just melt the complete ice-

The correct option is A 80^∘ C The mass of ice is equal to that of water #160; #160; i.e #160; #160; M_i = M_w = M Heat required to melt th ...

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Byju's Answer

Standard XII

Chemistry

Internal Energy

Take Cice =...

Question

Take $C_{i c e} = 0.53 c a l / g -^{\circ} C$ , $C_{w a t e r} = 1.0 c a l / g -^{\circ} C$ , $(L_{f})_{w a t e r} = 80 c a l / g$ and $(L_{v})_{w a t e r} = 529 c a l / g$ . Equal masses of ice (at $0^{\circ} C$ ) and water are in contact. Find the temperature of water needed to just melt the complete ice.

80^{\circ} C

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40^{\circ} C

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20^{\circ} C

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160^{\circ} C

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Solution

The correct option is A $80^{\circ} C$
The mass of ice is equal to that of water i.e $M_{i} = M_{w} = M$
Heat required to melt the complete ice $H_{R} = M_{i} (L_{f})_{w a t e r} = 80 M$ .......(1)
Let the temperature of the water be $T$
$∴$ Heat given by water to melt the ice $H_{G} = M_{w} C_{w} Δ T = M_{w} C_{w} (T - 0)$
$∴$ $H_{G} = M \times 1 \times T = M T$
But $H_{G} = H_{R}$ $⟹ M T = 80 M$
$⟹$ $T = 80^{o} C$

Suggest Corrections

Similar questions

Q. Take

C_{i c e} = 0.53 c a l / g -^{\circ} C

C_{w a t e r} = 1.0 c a l / g -^{\circ} C

(L_{f})_{w a t e r} = 80 c a l / g

and

(L_{v})_{w a t e r} = 529 c a l / g

. In a container of negligible mass

140 g

of ice initially at

- 15^{\circ} C

is added to

200 g

of water that has a temperature of

40^{\circ} C

. If no heat is lost to the surroundings, what is the final temperature of system in

^{\circ} C

Q. Take

C_{i c e} = 0.53 c a l / g -^{\circ} C

C_{w a t e r} = 1.0 c a l / g -^{\circ} C

(L_{f})_{w a t e r} = 80 c a l / g

and

(L_{v})_{w a t e r} = 529 c a l / g

. A nuclear power plant generates

500 M W

of waste heat that must be carried away by water pumped from a lake. If the water temperature is to rise

10^{\circ} C

, what is the required flow rate in

k g / s

Q. Take

C_{i c e} = 0.53 c a l / g -^{\circ} C

C_{w a t e r} = 1.0 c a l / g -^{\circ} C

(L_{f})_{w a t e r} = 80 c a l / g

and

(L_{v})_{w a t e r} = 529 c a l / g

. The temperatures of equal masses of three different liquids

A

B

and

C

are

12^{\circ} C

19^{\circ} C

and

28^{\circ} C

respectively. The temperature when

A

and

B

are mixed is

16^{\circ} C

and when

B

and

C

are mixed is

23^{\circ} C

. What would be the temperature when

A

and

C

are mixed ?

Q. Take

C_{i c e} = 0.53 c a l / g -^{\circ} C

C_{w a t e r} = 1.0 c a l / g -^{\circ} C

(L_{f})_{w a t e r} = 80 c a l / g

and

(L_{v})_{w a t e r} = 529 c a l / g

. A closely thermally. Insulated vessel contains

100 g

of water at

0^{\circ} C

. If the air from this vessel is rapidly pumped out, intensive evaporation will produce cooling and as a result of this water will freeze. How much ice will be formed by this method (in

g

) ? If latent heat of fusion is

80 c a l / g

and of evaporation

560 c a l / g

Q. When 40 g of ice at

0^{\circ} C

is added to 160 g of water at

40^{\circ} C

, find the final temperature?

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Take Cice=0.53cal/g−∘C, Cwater=1.0cal/g−∘C, (Lf)water=80cal/g and (Lv)water=529cal/g. Equal masses of ice (at 0∘C) and water are in contact. Find the temperature of water needed to just melt the complete ice.

The correct option is A 80∘CThe mass of ice is equal to that of water i.e Mi=Mw=MHeat required to melt the complete ice HR=Mi(Lf)water=80M .......(1) Let the temperature of the water be T ∴ Heat given by water to melt the ice HG=MwCwΔT=MwCw(T−0)∴ HG=M×1×T=MTBut HG=HR ⟹MT=80M⟹ T=80oC