What will be the final temperature when 150 g of ice at 0- C is mixed with 300 g of water at 50- C- Specific heat of water -1 cal-g- C- Latent heat of fusion of ice -80 cal-g-

The correct option is C T=6.7^∘ C Let us assume that T gt;0^∘ C Heat lost by water=heat gained by ice to melt + heat gained by water formed f ...

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Byju's Answer

Standard IX

Physics

Change of State

What will be ...

Question

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 $= 80 c a l / g$ .

T = {5.8}^{\circ} C

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T = {5.3}^{\circ} C

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T = {6.7}^{\circ} C

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T = {3.3}^{\circ} C

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Solution

The correct option is C $T = {6.7}^{\circ} C$
Let us assume that $T > 0^{\circ} C$
$H e a t l o s t b y w a t e r = h e a t g a i n e d b y i c e t o m e l t + h e a t g a i n e d b y w a t e r f o r m e d f r o m i c e$
$⟹ 300 \times 1 \times (50 - T) = 150 \times 80 \times + 150 \times 1 \times (T - 0)$
$⟹ T = {6.7}^{\circ} C$
Hence our assumption that $T > 0^{\circ} C$ is correct.

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Similar questions

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 = $80 c a l / g m$

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

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

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. If

10

gram of ice at

0^{\circ} C

is mixed with

10

gram of water at

40^{\circ} C

. The final mass of water in mixture is
(Latent heat of fusion of ice

= 80 c a l / g m

; specific heat of water

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

)

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What will be the final temperature when 150g of ice at 0∘C is mixed with 300g of water at 50∘C? Specific heat of water =1cal/g∘C. Latent heat of fusion of ice =80cal/g.

The correct option is C T=6.7∘CLet us assume that T>0∘CHeat lost by water=heat gained by ice to melt+heat gained by water formed from ice⟹300×1×(50−T)=150×80×+150×1×(T−0)⟹T=6.7∘CHence our assumption that T>0∘C is correct.

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 $= 80 c a l / g$ .