How much amount?

Classical Mechanics Level 4

100 g 100\text{ g} of solid A A of specific heat capacity 50 J/g ∘ C 50\text{ J/g}\, ^\circ\text{C} at 90 ∘ C 90\ ^\circ\text{C} is placed in a superconducting vessel of mass 10 g 10\text { g} of specific heat capacity 1 J/g ∘ C 1\text { J/g}\, ^\circ\text{C} . Then some amount of water at 40 ∘ C 40\ ^\circ\text{C} is added to the vessel to cool the mixture to 55 ∘ C 55\ ^\circ\text{C} .

Now, the above mixture after cooling down obtains a specific heat capacity of 25 J/g ∘ C 25\text{ J/g}\, ^\circ\text{C} . Then, to the above mixture is added a solid B B of specific heat capacity 20 J/g ∘ C 20 \text{ J/g}\, ^\circ\text{C} at 10 ∘ C 10\ ^\circ\text{C} such that the final temperature of the overall combination becomes 30 ∘ C 30\ ^\circ\text{C} .

How much amount of solid B B is added (up to 2 decimal places)?


The answer is 4520.83.

Ashish Menon by Ashish Menon , 20, India

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1 solution

Ashish Menon
Apr 9, 2016

Relevant wiki: Heat Transfer

Let the amount of water taken be x x g.
Amount of heat gained by water = m × c × θ R = x × 4.2 × 15 m × c × {\theta}_R = x × 4.2 × 15
Amount of heat lost by solid-1 = m × c × θ F = 100 × 50 × 35 m × c × {\theta}_F = 100 × 50 × 35
Amount of heat lost by vessel = m × c × θ F = 10 × 1 × 35 m × c × {\theta}_F = 10 × 1 × 35 (The vessel is superconducting so its initial temperature is equal to the initial temperature of the substance contained in it (i.e. solid-1)


According to law of conservation of energy,
Heat lost = Heat gained
∴ x × 4.2 × 15 = ( 100 × 50 × 35 ) + ( 10 × 1 × 35 ) ⟹ x = ( 175000 + 350 ) × 15 15 × 42 = 8350 3 \therefore x × 4.2 × 15 = (100 × 50 × 35) + (10 × 1 × 35)\\ \implies x = \dfrac{(175000 + 350) × 15}{15 × 42}\\ = \dfrac{8350}{3}

Let mass of solid-2 be y y g. For new mixture i.e. solid-1 + water + vessel, specific heat capacity = 25 J / g ° C 25J/g°C
Total mass = 8350 3 + 100 + 10 = 8680 3 \dfrac{8350}{3} + 100 + 10 = \dfrac{8680}{3}
Heat lost by mixture = m × c × θ F = 8680 3 × 25 × 25 m × c × {\theta}_F = \dfrac {8680}{3} × 25 × 25
Heat gained by solid-2 = m × c × θ R = y × 20 × 20 m × c × {\theta}_R = y × 20 × 20

According to law of conservation of energy,
Heat lost = Heat gained
∴ 8680 3 × 25 × 25 = y × 20 × 20 ⟹ y = 8680 × 25 × 25 3 × 20 × 20 ⟹ y = 27125 6 ⟹ y = 4520.83 g \therefore \dfrac{8680}{3} × 25 × 25 = y × 20 × 20\\ \implies y = \dfrac{8680 × 25 × 25}{3 × 20 × 20}\\ \implies y = \dfrac{27125}{6}\\ \implies y = \boxed{4520.83}g .

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Moderator note:

This solution is a miracle of bookkeeping.

Again no place to give my solution ?? So I give it here.
First heat exchange with W g of water :-
(100 \times 50 + 10 \times 1) \times (90 - 55)=W \times 4.186 \times (55- 40).
So W= 2793 g. Heat excgange with X g.of solid-2:-
(100 +10 +2793) \times 25 \times (55 - 30) = X \times 20 \times (30 -10).
So X=4535.9375 g. I have taken sp. heat of water as 4.186 since it was not given.

Niranjan Khanderia - 5 years, 1 month ago

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