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Calorimeter

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 $(L_{v})_{w a t e r} = 529 c a l / g$ . 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 $^{\circ} C$ ?

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Solution

Energy required to bring the ice to $0^{0} C$ and melt the ice completely is
$140 (0.53) (15) + 140 (80) = 12313 c a l$
Also, energy lost by water in cooling to $0^{0} C$ is
$200 (1) (40) = 8000 c a l$
Thus energy lost by water is not enough to melt the entire ice. Hence the ice and liquid phase will remain at equilibrium at $0^{0} C$ with partial ice melted.
Answer is $0^{0} C$ .

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Similar questions

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

(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.

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

(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

?

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

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

. The temperatures of equal masses of three different liquids

A

B

C

12^{\circ} C

19^{\circ} C

28^{\circ} C

respectively. The temperature when

A

B

16^{\circ} C

B

C

23^{\circ} C

. What would be the temperature when

A

C

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

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

. A closely thermally. Insulated vessel contains

100 g

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. Figure shows the temperature variation when heat is added continuously to a specimen of ice

(10 g)

- 40^{\circ} C

at constant rate. (Specific heat of ice

= 0.53 {cal/g}^{\circ} C

L_{ice} = 80 cal/g, L_{water} = 540 cal/g)

\begin{matrix} Column - I & Column - II (A) Value of Q_{1} (in cal) & (P) 800 (B) Value of Q_{2} (in cal) & (Q) 1000 (C) Value of Q_{3} (in cal) & (R) 5400 (D) Value of Q_{4} (in cal) & (S) 212 (T) 900 \end{matrix}

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