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Standard X

Physics

Principle of Calorimetry

1 kg of water...

Question

1 kg of water at $20^{\circ} C$ is, mixed with 800 g of water at $80^{\circ} C$ . Assuming that no heat is lost to the surroundings. Calculate the final temperature of the mixture.

{24.44}^{\circ} C

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{46.67}^{\circ} C

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{44.44}^{\circ} C

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{54.44}^{\circ} C

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Solution

The correct option is B ${46.67}^{\circ} C$

$m c θ_{1} + m c θ_{2} = m c θ_{0}$
since c is constant, assuming water is heated from $0^{o} C$
(1000)(20) + (800)(80)= (1800) $θ_{0}$
20000+64000= 84000=1800 $θ_{0}$
$θ_{0} = {46.67}^{o} C$
hence the finale temperature in $θ_{0} = {46.67}^{o}$

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Similar questions

1 k g

of water at

20^{o} C

is, mixed with

800 g

of water at

80^{o} C

. Assuming that no heat is lost to the surroundings, calculate the final temperature of the mixture :

10 g

of ice at

- 20^{\circ} C

is added to

20 g

of water at

20^{\circ} C

. Assuming that no heat is lost to the surrounding, find the final temperature of the mixture.
[Take, specific heat capacity of ice

C_{i} = 2100 {J kg}^{- 1}^{\circ} C^{- 1}

, specific heat capacity of water

C_{w} = 4200 {J kg}^{- 1}^{\circ} C^{- 1}

& Latent heat of fusion of ice

L = 3 .36 \times 10^{5} {J kg}^{- 1}

]

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

10 g

of ice at

- 20^{\circ} C

is added to

20 g

of water at

20^{\circ} C

. Assuming that no heat is lost to the surrounding, find the temperature of the mixture.
[Take, specific heat capacity of ice

C_{i} = 2100 {J kg}^{- 1}^{\circ} C^{- 1}

, specific heat capacity of water

C_{w} = 4200 {J kg}^{- 1}^{\circ} C^{- 1}

& Latent heat of fusion of ice

L = 3 .36 \times 10^{5} {J kg}^{- 1}

]

100 g

of ice (latent heat

80 {cal g}^{- 1}

), at

0^{\circ} C

is mixed with

100 g

of water (specific heat

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

) at

80^{\circ} C

. The final temperature of the mixture will be

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1 kg of water at 20∘C is, mixed with 800 g of water at 80∘C. Assuming that no heat is lost to the surroundings. Calculate the final temperature of the mixture.

The correct option is B 46.67∘Cmcθ1+mcθ2=mcθ0since c is constant, assuming water is heated from 0oC(1000)(20) + (800)(80)= (1800)θ020000+64000= 84000=1800θ0θ0=46.67oChence the finale temperature in θ0=46.67o

1 kg of water at $20^{\circ} C$ is, mixed with 800 g of water at $80^{\circ} C$ . Assuming that no heat is lost to the surroundings. Calculate the final temperature of the mixture.

The correct option is B ${46.67}^{\circ} C$

$m c θ_{1} + m c θ_{2} = m c θ_{0}$
since c is constant, assuming water is heated from $0^{o} C$
(1000)(20) + (800)(80)= (1800) $θ_{0}$
20000+64000= 84000=1800 $θ_{0}$
$θ_{0} = {46.67}^{o} C$
hence the finale temperature in $θ_{0} = {46.67}^{o}$