Question

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Physics Question, appeared in GCSE too!
Prove that when 'x' grams of steam at 100°C is mixed with 'y' grams of ice at 0°C and allowed to attain thermal equilibrium, the final temperature of mixture is $\sf{\theta = \dfrac{80(8x-y)}{x-y}}$.
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Answer 1

Correct Question:

Prove that when 'x' grams of steam at 100°C is mixed with 'y' grams of ice at 0°C and allowed to attain thermal equilibrium, the final temperature of mixture is $\sf{\theta = {\red{\bold{\dfrac{80(8x-y)}{x+y}}}}}$

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Given:

• Mass of steam= x gram
• Mass of ice = y gram
• Temperature of steam = 100°C
• Temperature of ice = 0°C
• Attained thermal equilibrium!

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Need to Prove:

Final temperature = $\dfrac{80(8x - y)}{(x + y)}$

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Let,

• L= Latent heat
• Cp= Specific heat
• Q1 = heat lost by stem
• Q2= heat gained by ice
• T= Final temperature

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Solution:

The amount of heat lost when the stem cools is:

$Q_{1}= x L_{vapour} + x Cp (100 -T)$

Put the values!

$\purple{\boxed{\bold{Q_{1} = 2254x + 4.18 x (100 - T)--i}}}$

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The amount of heat gained when the ice melts is:

$Q_{2} = y L_{freeze} + yCp (T -0)$

Put values!

$\purple{\boxed{\bold{Q_{2} = 334y + 4.18y T--ii}}}$

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We know,

Heat lost by the steam = Heat gained by the ice.

Equalizing equation i and ii we get,

$2254x + 4.18x (100 - T) = 334y + 4.18y T$

$\implies 2254x + 418x - 4.18x T = 334y + 4.18y T$

$\implies 2672x - 4.18x T = 334y+ 4.18y T$

$\implies 2672x - 334y = 4.18x T + 4.18y T$

$\implies 334 (8x-y) = 4.18 T (x + y)$

$\red{\large{\bold{\boxed{\implies T = \dfrac{80(8x-y)}{x + y}}}}}$

Answer 2

$\huge\bf{\underline{\color{indigo}GIVEN}}$

• $\bf\red{Amount~of~steam~}$ = x grams

• $\bf\red{Amount~of~ice~}$ = y grams

$\bf\:~~~~~\circ$ 'x' grams of steam at 100°C is mixed with 'y' grams of ice at 0°C .

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$\huge\bf{\underline{\color{orange}CONCEPT}}$

$\mathbb{\underline{LAW\:OF\:MIXTURES\::\:\atop{PRINCIPLE\:OF\:CALORIMETRY\:}}}$

❍ When two bodies at different temperatures are mixed, heat will be transferred from body at higher temperature to a body at lower temperature till both acquire same temperature . The body at higher temperature releases heat while body at lower temperature absorbs it, so that

$\pink\bullet\:\bf{\boxed{\color{mediumspringgreen}HEAT~LOST~=~HEAT~GAINED~}}$

✯ Principle of Calorimetry represents the law of conservation of heat energy .

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$\\ \huge\bf{\underline{\color{olive} SOLUTION}}$

Let,

• $\bf\blue{Final\: temperature\:=\:\theta\:}$

CASE - 1 ;-

ᗕᗒ Heat required by ice to raised it's temperature to θ°C,

$\bf{{\boxed{0°C~(ice)}}\:\underset{Q_1}{\longrightarrow}\:{\boxed{0°C~(water)}}\:\underset{Q_2}{\longrightarrow}\:{\boxed{{\theta}°C~(water)}}\:}$

$\bigstar\:\bf\gray{Q_1\:=\:m_{ice}\:L_{ice}\:}$

Where,

• $\bf{\red{L_{ice}~=~Latent~heat~of~fusion~for~ice}}~\atop{=~80~ cal/g}$

$\rm{:\implies\:Q_1\:=\:y\times{80}\:}$

$\bf{:\implies\:Q_1\:=\:80y\:Cal}$

And

$\bigstar\:\bf\gray{Q_2\:=\:m_{ice}\:C_{water}\:\triangle{\theta}\:}$

Where,

• $\bf{\red{C_{water}\:=\: Specific\:heat\:of\:water}}~\atop{=~1 ~cal/g.°C}$

• $\bf\red{\triangle{\theta}\:=\: Change\:in\: temperature\:}$

$\rm{:\implies\:Q_2\:=\:y\times{1}\times{(\theta\:-\:0)}\:}$

$\rm{:\implies\:Q_2\:=\:y\times{\theta}\:}$

$\bf{:\implies\:Q_2\:=\:y{\theta}\:Cal}$

So total heat(Q) required by ice is,

➝ $\bf{Q_{ice}\:=\:Q_1\:+\:Q_2\:}$

➝ $\bf\green{Q_{ice}\:=\:(80y\:+\:y{\theta})\:Cal\:}$

CASE - 2 ;-

ᗕᗒ Heat given by steam when condensed,

$\bf{{\boxed{100°C~(steam)}}\:\underset{Q_1}{\longrightarrow}\:{\boxed{100°C~(water)}}\:\underset{Q_2}{\longrightarrow}\:{\boxed{{\theta}°C~(water)}}\:}$

$\bigstar\:\bf\gray{Q_1\:=\:m_{steam}\:L_{water}\:}$

Where,

• $\bf{\red{L_{water}~=~Latent~heat~of~water}}~\atop{=~540~ cal/g}$

$\rm{:\implies\:Q_1\:=\:x\times{540}\:}$

$\bf{:\implies\:Q_1\:=\:540x\:Cal}$

And

$\bigstar\:\bf\gray{Q_2\:=\:m_{steam}\:C_{water}\:\triangle{\theta}\:}$

$\rm{:\implies\:Q_2\:=\:x\times{1}\times{(100\:-\:{\theta})}\:}$

$\rm{:\implies\:Q_2\:=\:x\times{(100\:-\:{\theta})}\:}$

$\bf{:\implies\:Q_2\:=\:(100x\:-\:x{\theta})\:Cal}$

So total heat(Q) given by steam is,

➝ $\bf{Q_{steam}\:=\:Q_1\:+\:Q_2\:}$

➝ $\rm{Q_{steam}\:=\:540x\:+\:(100x\:-\:x{\theta})\:}$

➝ $\bf\green{Q_{steam}\:=\:(640x\:-\:x{\theta})\:Cal}$

☯︎ Now, according to the Principle of Calorimetry,

$\huge\red\checkmark$ $\bf{Q_{steam}\:=\:Q_{ice}\:}$

$\longmapsto\:\rm{640x\:-\:x{\theta}\:=\:80y\:+\:y{\theta}\:}$

$\longmapsto\:\rm{640x\:-\:80y\:=\:y{\theta}\:+\:x{\theta}\:}$

$\longmapsto\:\rm{(y\:+\:x)\:{\theta}\:=\:80\:(8x\:-\:y)}$

$\longmapsto\:\rm{{\theta}\:=\:\dfrac{80\:(8x\:-\:y)}{y\:+\:x}\:}$

$\pink\longmapsto\:\bf{\purple{{\theta}\:=\:\dfrac{80\:(8x\:-\:y)}{x\:+\:y}\:}}~~~~{(Hence~proved)}$