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

Byju's Answer

Standard XII

Physics

Equipartition Theorem

The water of ...

Question

The water of mass $75 g$ at $100^{o} C$ is added to ice of mass $20 g$ at ${- 15}^{o} C$ What is the resultIng temperaure. (Latent heat of $i c e = 80 \frac{c a l}{g}$ and specific heat of $i c e = 05$ ) approx?

0^{o}

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

20^{o}

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

60^{o}

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

100^{o} C

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

Open in App

Solution

The correct option is C $60^{o}$
Let final temperature is T
Then,
$H e a t l o s s = H e a t g a i n$
$75 \times 1 \times (100 - T) = (20 \times 0.5 \times 15) + (20 \times 80) + (20 \times 1 \times (T - 0))$
$5750 = 95 T$
$T = {60.5}^{o} C$

Suggest Corrections

Similar questions

Q. The water of mass

75

g at

100^{\circ} C

is added to ice of mass

20

g at

- 15^{\circ} C

. What is the resulting temperature approximately? (Latent heat of ice

= 80 \frac{c a l}{g}

and Specific Heat of ice

= 0.5

)

500 g

of water and

100 g

of ice at

0^{o} C

are in a calorimeter whose water equivalent is

40 g

10 g

of steam at

100^{o} C

added to it. Then water in the calorimeter is (Latent heat of ice

= 80 c a l / g

, Latent heat of steam

= 540 c a l / g

)

Q. How much heat is required to change

10 g

ice at

0^{o} C

to steam at

100^{o} C

? Latent heat of fusion and vaporization for

H_{2} O

are

80 c a l / g

and

540 c a l / g

respectively. Specific heat of water is

1 c a l / g

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

]

1 g

of steam at

100^{o} C

and equal mass of ice at

0^{o} C

are mixed. The temperature of the mixture in steady state will be (latent heat of steam

= 540 c a l / g

, latent heat of ice

= 80 c a l / g

Join BYJU'S Learning Program

The water of mass 75g at 100oC is added to ice of mass 20g at −15oC What is the resultIng temperaure. (Latent heat of ice=80calg and specific heat of ice=05) approx?

The correct option is C 60oLet final temperature is TThen,Heat loss=Heat gain75×1×(100−T)=(20×0.5×15)+(20×80)+(20×1×(T−0))5750=95TT=60.5oC

The water of mass $75 g$ at $100^{o} C$ is added to ice of mass $20 g$ at ${- 15}^{o} C$ What is the resultIng temperaure. (Latent heat of $i c e = 80 \frac{c a l}{g}$ and specific heat of $i c e = 05$ ) approx?

The correct option is C $60^{o}$
Let final temperature is T
Then,
$H e a t l o s s = H e a t g a i n$
$75 \times 1 \times (100 - T) = (20 \times 0.5 \times 15) + (20 \times 80) + (20 \times 1 \times (T - 0))$
$5750 = 95 T$
$T = {60.5}^{o} C$