The water of mass 75g at 100-C is added to ice of mass 20 g at -15-C- What is the resulting temperature approximately- Latent heat of ice -80 cal g and Specific Heat of ice -0-5

The correct option is B 60^∘C Let final temperature is T . #10; #10; #160; #10; #10;Then, #10; #10;Heat Loss = Heat Gain #10; ...

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Byju's Answer

Standard VII

Chemistry

Latent Heat of Vaporisation

The water of ...

Question

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 $= 0.5$ )

0^{\circ} C

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20^{\circ} C

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60^{\circ} C

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100^{\circ} C

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Solution

The correct option is B $60^{\circ} C$
Let final temperature is $T$ .

Then,

Heat Loss $=$ Heat Gain

$75 \times 1 \times (100 - T) = 20 \times 0.5 \times 15 + 20 \times 80 + 20 \times 1 \times (T - 0)$

$7500 - 75 T = 150 + 1600 + 20 T$

$95 T = 5750$

$T = 60.5^{\circ} C \approx 60^{\circ} C$

Hence, this is the required result.

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Similar questions

Q. The water of mass

75 g

100^{o} C

is added to ice of mass

20 g

{- 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?

Q. Ice at

0^{\circ} C

is added to

1 g

of steam at

100^{\circ} C

. For

x g

of ice added, the temperature of steam reduces to

0^{\circ} C

. Find the value of

x

.
(Latent heat of ice

= 80 cal/g

and latent heat of steam

= 540 cal/g

. Also, specific heat of water

= 1 cal/ kg K

)

1 g

of steam at

100^{\circ} C

and an equal mass of ice at

0^{\circ} 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

, specific heat of water

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

)

Q. The amount of heat required to convert

1

g of ice at

- 10^{0} C

into steam at

100^{0} C

is (approximately): (specific heat of water

=

4.187

J/g

^{0}

C, latent heat of fusion

=

334

J/g , latent heat of vaporization

=

2260

J/g, specific heat of ice

=

2.03

J/g) :

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

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The water of mass 75g at 100∘C is added to ice of mass 20 g at −15∘C. What is the resulting temperature approximately? (Latent heat of ice =80 calg and Specific Heat of ice =0.5)

The correct option is B 60∘CLet final temperature is T. Then, Heat Loss = Heat Gain 75×1×(100−T)=20×0.5×15+20×80+20×1×(T−0) 7500−75T=150+1600+20T 95T=5750 T=60.5∘C≈60∘C Hence, this is the required result.

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 $= 0.5$ )