Question

A mass of 50 g of a certain metal at 150°C is immersed in 100 g of water at 11°C. The final temperature is 20°C. Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is :-
$\large\underline{\sf{ \rm \: 4.2 \: joule \: {g}^{ - 1} \: {k}^{ - 1} }}$
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Answer 1

Answer:

Given :-

• A mass of 50 g of a certain metal at 150°C is immersed in 100 g of water at 11°C.
• The final temperature is 20°C.
• The specific heat capacity of water is 4.2 J g-¹ k-¹.

To Find :-

• What is the specific heat capacity of the metal.

Solution :-

First, we have to convert :

✫ Initial temperature of metal :

$\leadsto \sf Initial\: Temperature_{(Metal)} =\: 150^{\circ}C$

$\leadsto \sf Initial\: Temperature_{(Metal)} =\: 150^{\circ}C + 273$

$\leadsto \sf\bold{\blue{Initial\: Temperature_{(Metal)} =\: 423\: K}}$

✫ Final Temperature :

$\leadsto \sf Final\: Temperature =\: 20^{\circ}C$

$\leadsto \sf Final\: Temperature =\: 20^{\circ}C + 273$

$\leadsto \sf\bold{\orange{Final\: Temperature =\: 293\: K}}$

✫ Initial Temperature of water :

$\implies \sf Initial\: Temperature_{(Water)} =\: 11^{\circ}C$

$\implies \sf Initial\: Temperature_{(Water)} =\: 11^{\circ}C + 273$

$\implies \sf\bold{\green{Initial\: Temperature_{(Water)} =\: 284\: K}}$

Now, we have to find the specific heat capacity of the metal :

Let,

$\small \mapsto \bf Specific\: Heat\: Capacity_{(Metal)} =\: c_{(metal)}$

Given :

◉ Mass of metal ($\sf m_{[metal]}$) = 50 g

◉ Mass of water ($\sf m_{[water]}$) = 100 g

◉ Specific Heat Capacity of water ($\sf c_{[water]}$) = 4.2 J g-¹ k-¹

◉ Initial Temperature of Metal ($\sf T_{[initial]}$) = 423 K

◉ Final Temperature ($\sf T_{[Final]}$) = 293 K

◉ Initial Temperature of Water ($\sf T_{[Initial]}$) = 284 K

According to the question by using the formula we get,

$\small \implies \bf Heat\: lost\: by\: Metal =\: Heat\: gained\: by\: Water$

As we know that :

$\small \bigstar \: \: \sf\boxed{\bold{\pink{m_{(metal)} c_{(metal)} \triangle T_{(metal)} =\: m_{(water)} c_{(water)} \triangle T_{(water)}}}}\: \: \: \bigstar$

where,

• $\sf m_{(metal)}$ = Mass of metal
• $\sf c_{(metal)}$ = Specific heat capacity of metal
• $\sf \triangle T_{(metal)}$ = Change of temperature of metal
• $\sf m_{(water)}$ = Mass of water
• $\sf c_{(water)}$ = Specific heat capacity of water
• $\sf \triangle T_{(water)}$ = Change of temperature of water

So, now by putting those values we get :

$\small \implies \bf m_{(metal)} c_{(metal)} \triangle T_{(metal)} =\: m_{(water)} c_{(water)} \triangle T_{(water)}$

$\implies \sf 50 \times c_{(metal)} \times (423 - 293) =\: 100 \times 4.2 \times (293 - 284)$

$\implies \sf 50 \times c_{(metal)} \times 130 =\: 420 \times 9$

$\implies \sf 50 \times 130 \times c_{(metal)} =\: 3780$

$\implies \sf 6500 \times c_{(metal)} =\: 3780$

$\implies \sf c_{(metal)} =\: \dfrac{378\cancel{0}}{650\cancel{0}}$

$\implies \sf c_{(metal)} =\: \dfrac{378}{650}$

$\implies \sf\bold{\red{c_{(metal)} =\: 0.582\: J\: g^{- 1}\: k^{- 1}}}$

$\small \sf\bold{\purple{\underline{\therefore\: The\: specific\: heat\: capacity\: of\: metal\: is\: 0.582\: J\: g^{- 1}\: k^{- 1}\: .}}}$

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Answer 2

Answer:

Let subscript 1 denote solid (hot body) and subscript 2 denote water (cold body)

Initial temp. of solid T₁=150°C

Mass of solid m₁ = 50 g = 0.05 Kg

Initial temp. of water T₂=11°C

Mass of water m₂= 100 g = 0.1 Kg

Final temp of both T= 20°C

Specific heat capacity of water c₂=4.2 J/g°C

Let Specific heat capacity of solid be c₁

now,

heat gained by water = heat lost by solid

heat gained/lost = mcΔT

∴ m₁c₁ΔT₁ = m₂c₂ΔT₂

⇒ 0.05*c₁*(150-20) = 0.1*4.2*(20-11)

⇒ 0.05*130*c₁ = 0.1*4.2*9

⇒ c₁ = 0.5815 J/g°C

Answer: Specific heat capacity of the solid is 0.5815 J/g°C