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Melting and Boiling point

nt10 gms of i...

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nt10 gms of ice at -10 deg c is mixed with 100 gm of water at 50 deg c contained in a calorimetr weighing 50 gm. (Specific heat of water = 1 cal/gm, Latent heat = 80 cal/gm, specific heat of ice = 0.5 cal/gam/deg c and the specific heat of copper = 0.09 cal/gm/deg c. The final temperature reached by the mixturen

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50 g m

10^{0} C

20 g m

100^{0} C .

When the mixture finally reaches its steady state inside a calorimeter of water equivalent

1.5 g m

then : [Assume calorimeter was initially at

0^{0} C,

Take latent heat of vaporization of water

= 540

cal/gm, Latent heat of fusion of water

= 80

cal/gm, specific heat capacity of water

= 1 c a l / g m -^{0} C,

specific heat capacity of ice

= {0.5 cal/ gm}^{0} C

]

5 g m

30^{\circ} C

5 g m

- 20^{\circ} C

are mixed together in a calorimeter. Find the final temperature of mixture. Water equivalent of calorimeter is negligible, specific heat of ice

= 0.5 c a l / g m^{\circ} C

and latent heat of ice

= 80 c a l / g m

20 gm

- 10^{\circ} C

m gm

100^{\circ} C

. The minimum value of

m

so that finally all ice and steam converts into water at

0^{\circ} C

(

Use: specific heat and latent heat as

C_{i c e} = 0.5 cal/gm^{\circ} C, C_{water} = 1 cal/gm^{\circ} C, L_{f u s i o n}

= 80 cal/gm

L_{v a p o r}

= 540 cal/gm

)

Q. The amount of heat required to convert 1 g of ice (specific 0.5 cal at

g^{- 1 o} C^{- 1}

- 10^{0} C

100

^{\circ} C

is ___________.

[ Given: Latent heat of ice is

80 C a l / g m,

Latent heat of steam is

540 C a l / g m

, Specific heat of water is

1 C a l / g m / C

What will be the final temperature, when $150 g m$ of ice at $0^{\circ} C$ is mixed with $300 g m$ of water at $50^{\circ} C$ . Specific heat of water = $1 c a l / g m /^{\circ} C$ . Latent heat of fusion of ice = $80 c a l / g m$

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