You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Physics

Latent Heat of Fusion

Find the amou...

Question

Find the amount of heat in calories required to convert 10g of ice at –10°C to water at 10°C

[C of ice = 0.5 cal/g°C, C of water = 4.2J/g°C = 1 cal/g°C, L of ice = 80cal/g]

95 J

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

950 calories

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

100 J

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

1000 calories

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

Open in App

Solution

The correct option is B
950 calories

Suggest Corrections

Similar questions

1 kg

of ice at

- 10^{\circ} C

is mixed with

4.4 kg

of water at

30^{\circ} C

. The final temperature of mixture is

Given specific heat of ice

= 0.5 {cal/g}^{\circ} C

,
Specific heat of water

= 1 {cal/g}^{\circ} C

,
Latent heat of fusion for ice

= 80 cal/g

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

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

) at

- 10^{0} C

to steam at

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

]

Q. Calculate the amount of heat (in calories) required to concert 5 g m ice at

0^{\circ} C

to steam at

100^{\circ}

Q. The amount of heat (in calories) required to convert 10g of ice at -

10^{0} C

into steam at

100^{0} C

is nearly:

(Specific heat of steam is

536 c a l / g m - K

)

Q. In a process, 10 gm of ice at

- 5^{o} C

is converted into steam at

100^{o} C

.If latent heat of fusion of ice is 80 cal

g^{- 1}

, then the amount of heat required to convert 10 g of ice at

0^{\circ} C

into 10 g of water at same temperature is :

Join BYJU'S Learning Program

Find the amount of heat in calories required to convert 10g of ice at –10°C to water at 10°C [C of ice = 0.5 cal/g°C, C of water = 4.2J/g°C = 1 cal/g°C, L of ice = 80cal/g]

The correct option is B 950 calories

Find the amount of heat in calories required to convert 10g of ice at –10°C to water at 10°C

[C of ice = 0.5 cal/g°C, C of water = 4.2J/g°C = 1 cal/g°C, L of ice = 80cal/g]

The correct option is B
950 calories