You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

1st Law of Thermodynamics

150 g of ice ...

Question

$150 g$ of ice is mixed with $100 g$ of water at temperature $80^{o} C$ . The latent heal of ice is $80 c a l / g$ and the specific heat of water is $1 c a l / g^{o} C$ . Assuming no heat loss to the environment, the amount of ice which does not melt is :

100 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

150 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

50 g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $50 g$
Let the ice is at $0^{o} C$ .
The maximum heat given by water to ice will be:
$Q = m_{w} c_{w} Δ = 100 \times 1 \times (80 - 0) = 8000 c a l .$
Let mass $m$ of the ice is melt , by the heat given by water to ice.
Hence, $Q = m L$
$8000 = m \times 80$
$m = 8000 / 80 = 100 g$
Therefore, mass of ice which doesn't melt will be: $m^{'} = 150 - 100 = 50 g$

Suggest Corrections

Similar questions

Q. 150 g of ice is mixed with 100 g of water at temperature

80^{o} C

. The latent heat of ice is 80 cal/g and the specific heat of water is 1 cal/g

-^{o} C

. Assuming no heat loss to the environment, the amount of ice which does not melt is

Q. What will be the final temperature when

150 g

of ice at

0^{\circ} C

is mixed with

300 g

of water at

50^{\circ} C

? Specific heat of water

= 1 c a l / g^{\circ} C

. Latent heat of fusion of ice

= 80 c a l / g

Q. Ice in a freezer is at

- 7^{\circ} C . 100 g

of this ice is mixed with

200 g

of water at

15^{\circ} C

. Take the freezing temperature of water to be

0^{\circ} C

, the specific heat of ice equal to

2.2 J / g^{\circ} C

, specific heat of water equal to

4.2 J / g^{\circ} C

, and the latent heat of ice equal to

335 J / g

. Assuming no loss of heat to the environment, the mass of ice in the final mixture is closest to

Q. The amount of heat required to raise the temperature of

75 g

of ice at

0^{o} C

to water at

10^{o} C

is:

[latent heat of fusion of ice is

80 c a l / g

, specific heat of water is

1 c a l / g^{o} C

]

100 g

of ice (latent heat

80 c a l g^{- 1}

,at

0^{o} C

) is mixed with

100 g

of water (specific heat 1 cal

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

) at

80^{o} C

. The final temperature of the mixture will be:

Join BYJU'S Learning Program

150 g of ice is mixed with 100 g of water at temperature 80oC. The latent heal of ice is 80 cal/g and the specific heat of water is 1cal/goC. Assuming no heat loss to the environment, the amount of ice which does not melt is :

$150 g$ of ice is mixed with $100 g$ of water at temperature $80^{o} C$ . The latent heal of ice is $80 c a l / g$ and the specific heat of water is $1 c a l / g^{o} C$ . Assuming no heat loss to the environment, the amount of ice which does not melt is :