Heat gained is Heat lost

Q1

A solid of mass 56 g (Sp. heat capacity 0.9 J/kg K) and at 220°C is placed in 220 g of water at 20°C. Calculate final temperature of mixture?

Sol:

Substance	Mass	S.H.C.	Initial Temp	Final temp = x
Hot solid	56 g	0.9 J/kg K	220°C	θ_R = (x−20°C)
Cold water	220 g	4.2 J/kg K	20°C	θ_F = (220°C−x)

Heat given out by solid	=	mC θ_F
	=	56 × 0.9 × (220 − x)
	=	11088 − 50.4 x
Heat absorbed by cold water	=	mC θ_R
	=	220 × 4.2 × (x − 20)
	=	924x − 18480
Heat given out by solid	=	Heat absorbed by water
11088 − 50.4 x	=	924x − 18480
∴ 974.4x	=	29568
∴ x	=	30.34°C

Q2

55 g of hot water is poured in 125 g of cold water, when temperature of cold water rises by 10°C. If the temperature of hot water is 85°C, calculate the initial temperature of cold water?

Sol:

Substance	Mass	S.H.C.	Initial Temp	Final temp = x
Hot water	55 g	4.2Jg⁻¹°C⁻¹	85°C	θ_R = 10°C
Cold water	125 g	4.2Jg⁻¹°C⁻¹	?	θ_F = (85−x)

Heat given out by hot water	=	mC θ_F
	=	55 × 4.2 × (85 −x)
Heat absorbed by cold water	=	mC θ_R
	=	125 × 4.2 × 10
∴ Heat given out by hot water	=	Heat absorbed by cold water.
55 × 4.2 (85 − x)	=	125 × 4.2 × 10
19635 − 231 x	=	5250
231 x	=	19635 − 5250
∴ x	=	14385/231
Thus, temperature of mixture	=	62.27°C
∴ Initial temp, of cold water	=	(62.27 − 10)
	=	52.27°C.

Q3

A solid of mass 120 g and at 110°C, is dropped into 220 g of water at 10°C, which is contained in a copper calorimeter of mass 100 g. The final temperature attained by mixture is 20°C. If the specific heat capacity of copper and water is 390 J kg K^{− 1} and 4200 J kg^{− 1} K^{− 1}. Calculate the sp. heat capacity of solid in S.I. system ?

Sol:

Substance	Mass	S.H.C.	Initial Temp	Final Temp = 20°C
solid	l20g	?	110°C	θ_R = (20−10) = 10°
Water	220 g	4.2 Jg⁻¹°C⁻¹	10°C	θ_F = (110−20) = 90°C
Calorimeter	100g	0.39Jg⁻¹°C⁻¹	10°C	θ_F=(110−20) = 90°C

Heat lost by solid	=	mC θ_F
	=	120 × C × 90
	=	10800 C
Heat gained by water	=	mC θ_R
	=	220 × 4.2 × 10
	=	9240 J
Heat gained by Calorimeter	=	mC θ_R
	=	100 × 0.39 × 10
	=	390 J
∴ Total heat gained	=	9240 + 390
	=	9630 J
Heat lost	=	Heat gained
10800 × C	=	9630
∴ C	=	9630/10800
	=	0.892 Jg^{− 1}° C^{− 1}
	=	892 J kg K^{− 1}

Q4

120 g of water at 64°C, is poured into vessel containing 100 g of water at 22°C. The final temp, recorded is 31°C. Calculate thermal capacity of vessel?

Sol:

Substance	Mass	S.H.C.	Initial Temp	Final Temp = 31°C
Hot water	120g	4.2J/g°C	64°C	θ_F = (64−31) = 33° C
Cold water	100 g	4.2J/g°C	22°C	θ _R = (31 − 22)°C = 9° C
Vessel	?(m)	?(c)	22°C	θ_R = (31 − 22) = 9° C

Let thermal capacity of vessel	=	x
	=	mC
Heat lost by hot water	=	mC θ_F
	=	120 × 4.2 × 33
	=	16632 J
Heat gained by cold water	=	mC θ_R
	=	100 × 4.2 × 9
	=	3780 J
Heat gained by vessel	=	mC θ_R
	=	x × 9°C
Heat gained	=	Heat lost
3780 J + x × 9°C	=	16632 J
∴ x	=	(16632−3780)/9	J°C^{− 1}
	=	1428 J°C^{− 1}

Q5

A liquid P of specific heat capacity 1800 J kg^{− 1} K ^{− 1} and at 55°C, is mixed with liquid R of specific heat capacity 1060 J kg^{− 1} K^{− 1} and at 18°C. After mixing the temperature of mixture is 45°C. In what proportion by weight are the liquids mixed ?

Sol:

Substance	Mass	S.H.C.	Initial Temp	Final temp = 45°C
Liq. P	?	1800J kg^{− 1} K^{− 1}	55°C	θ_R = (45 − 18) = 27° C
Liq. R	?	1060 J kg^{− 1} K^{− 1}	18°	θ_F = (55 −45) = 10°C

Heat lost by liq. P	=	mC θ_F
	=	P × 1800 × 10
Heat gained by liq. R	=	mC θ_R
	=	R × 1060 × 27
Heat Gained	=	Heat lost
R × l060 × 27	=	P × 1800 × 10
∴ P/R	=	1060 × 27/1800 × 10
	=	28620/18000
	=	4.77/3
∴ P : R	=	4.77 : 3.

Q6

25 g of ice at − 20°C is heated by a burner supplying heat energy at a rate of 200 J/s. Calculate the time, in which the water formed from ice attains a temp. of 80°C. Specific heat capacity of ice is 2.1 J g^{− 1}°C^{− 1} and specific latent heat of ice Is 336 J g^{− 1}?

Sol:

Heat gained by ice to attain temp. of 0°C	=	mc θ_R
	=	25 × 2.1 × 20
	=	1050 J
Heat gained by ice to form water at 0C	=	mL°_ice
	=	25 × 336
	=	8400 J
Heat gained by water till 80°C	=	mc θ_R
	=	25 × 4.2 × 80
	=	8400 J
∴ Total heat gained	=	(1050 + 8400 + 8400) J
	=	17850 J
Also rate of supply of heat	=	200 J/s
∴ Time req.	=	17850/200
	=	89.25 s.

Q7

Calculate the amount of ice, which is sufficient to cool 180 g of water, contained in a vessel of mass 25 g (S.H.C. 0.5 Jg^{− 1}°C) at 40°C, such that final temp. of mixture is 10°C? [Specific latent heat of ice is 336 Jg^{− 1}]

Sol:

Substance	Mass	S.H.C/S.L.H	Initial Temp.	Final Temp. = 10°c
Ice	?(x)	336 Jg^{− 1}	0°	θ_R = (10 − 0) = 10°C
Water	180 g	4.2 Jg^{− 1}°C^{− 1}	40°C	θ_F = (40 − 10) = 30°C
Vessel	25 g	0.5 Jg^{− 1}°C^{− 1}	40°C	θ_F = (40 − 10) = 30°C

Heat gained by ice to form water at 0°C	=	mL
	=	x × 336
Heat gained by water formed from ice	=	mc θ_R
	=	x × 4.2 × 10
	=	42x
∴ Total heat gained	=	336 x + 42 x
	=	378 x
Heat lost by water at 40° C	=	mc θ_F
	=	180 × 4.2 × 30
	=	22680 J
Heat lost by vessel at 40° C	=	mc θ_F
	=	180 × 4.2 × 30
	=	22680 J
Heat lost by vessel at 40°C	=	mc θ_F
	=	25 × 0.5 × 30
	=	375 J
Total heat lost	=	(22680 + 375)
	=	23055 J
Heat gained	=	Heat lost
378 x	=	23055
∴ x (Mass of ice)	=	23055/378
	=	60.99 g.

Q8

A metal ball of mass 264 g and at 800°C is placed on the block of ice, when 512 g of ice melts. If the sp. heat capacity of metal ball is 0.8 Jg^{− 1}°C^{− 1}. Calculate sp. latent heat of ice ?

Sol:

As the block does not melt completely therefore final temperature is zero degree celsius.

Substance	Mass	S.H.C/S.L.H	Initial Temp.	Final Temp = 0°C
Ice	512 g	? (L)	0°C	∴ θ_R = 0°C
Metal ball	264 g	0.8 Jg^{− 1}°C^{− 1}	800°	∴ θ_F = (800 − 0) = 800°C

Heat gained by metal to melt	=	mL
	=	512 g x L
Heat lost by metal ball	=	mc θ_F
	=	264 × 0.8 × 800
	=	168960 J
Heat gained	=	Heat lost
512 g × L	=	168960 J
∴ L	=	168960/512
	=	330 Jg^{− 1}

Q9

15 kg of hot water in a bucket at 65°C is cooled by mixing 8 kg of water at 5°C. What is the final temperature of the mixture ? (Neglect the heat absorbed by the bucket)

Sol:

Suppose the final temperature (equilibrium temperature or resultant temperature) of the mixture be x°C.

We will now calculate the "heat lost by hot water" and the "heat gained by cold water" in terms of x.

(i) Calculation of Heat Lost by Hot Water
Mass of hot water, m₁	=	15 kg
Specific heat of hot water	=	C₁
Initial temperature of hot water	=	65°C
Final temperature of mixture	=	x°C
So, Fall in temp, of hot water, T₁	=	(65 − x)°
Now,Heat lost by hot water, Q₁	=	m₁ × C₁ × T₁
(or)	=	15 × C₁ × (65 − x) ----- (i)
(ii) Calculation of Heat Gained by Cold Water
Mass of cold water, m2	=	8 kg
Specific heat of cold water	=	C₂
Initial temperature of cold water	=	5°C
Final temperature of mixture	=	x°C
So, Rise in temp, of cold water, T₂	=	(x − 5)°
Now,Heat gained by cold water, Q₂	=	m₂ × C₂ × T₂
or	=	8 × C₂ × (x − 5)
We know that according to the principle of calorimetry :
Heat lost	=	Heat gained