QBManager

Thermal Properties of Matter

146648 If $60 %$ of the kinetic energy of water falling from $210 m$ high water fall is converted into heat. The raise in temperature of water at the bottom of the falls is nearly (specific heat of water $= 4.2 \times 10^{3} {Jkg}^{- 1} K^{- 1}$ )

1

0.6^{\circ} C

2

{0.3}^{\circ} C

3

1.2 K

4

2.4 K

Explanation:

B
Exp:
Given, $h = 210 m$
Specific heat of waters $= 4.2 \times 10^{3} J {kg}^{- 1} K^{- 1}$
As we know that,
When water falls on the surface of earth, then its potential energy is converted into kinetic energy.
$∴$ Kinetic energy $=$ Potential energy
$K . E = mgh$
According to question,
$mc Δ T = \frac{60}{100} \times mgh$
$Δ T = \frac{0.6 gh}{s} = \frac{0.6 \times 10 \times 210}{4.2 \times 10^{3}}$
$Δ T = {0.3}^{\circ} C$

AP EAMCET (18.09.2020) Shift-I

Thermal Properties of Matter

146649 The specific heat capacities of an ideal gas at the constant pressure and at constant volume are $620 {Jkg}^{- 1} k^{- 1}$ and $420 {Jkg}^{- 1} K^{- 1}$ Respectively. The density of the gas at STP is approximately,

1

2.88 {kgm}^{- 3}

2

4.86 {kgm}^{- 3}

3

3.88 {kgm}^{- 3}

4

1.86 {kgm}^{- 3}

Explanation:

D Given,
Specific heat capacity of a gas at constant pressure, $c_{p} =$ $620 {Jkg}^{- 1} K^{- 1}$
Specific heat capacity of gas at constant volume, $c_{V} =$ $420 {Jkg}^{- 1} K^{- 1}$
$∴$ Molar specific heat of gas at constant pressure and constant volume are given by
$∴ c_{p} = {mc}_{p} = m (620)$
$c_{v} = {mc}_{p} = m (420)$
As we know that,
$∴ c_{P} - c_{V} = R$ (Here, $R =$ universal gas constant)
$m (620 - 420) = R$
$m = \frac{R}{200}$
Where, $μ =$ number of molar of gas
$PV = RT (∴$ for 1 mole, $μ = 1)$
$Pm = ρ RT$
( $∴$ density, $ρ = \frac{m}{V}$ )
$10^{5} \times \frac{R}{200} = ρ R (273)$
${STP, P = 1 \times 10^{5} atm, T = 273 K}$
$ρ = 1.85 {kgm}^{- 3}$

AP EAMCET (23.04.2019) Shift-I

Thermal Properties of Matter

146650 Electrical energy costs 25 paisa per kilowatt hour. Assuming that no energy is wasted, the cost of heating $4.6 kg$ of water from $25^{\circ} C$ to the boiling point is

1 25 paisa

2 50 paisa

3 20 paisa

4 10 paisa

Explanation:

D Given,
Electric energy $cost (C) = 25$ paisa/ $KWh$
Mass of water, $m = 4.6 kg$
Initial temperature of water $T_{i} = 25^{\circ} C$
Final temperature of water $T_{f} = 100^{\circ} C$ (Boiling point of water) as heat used by water to boil,
$Q = ms (T_{f} - T_{i})$
So, $Q = 4.6 \times 4.2 \times 10^{3} \times (100 - 25)$
$Q = 1449 \times 10^{3} J (∴ s = 4.2 \times 10^{3} J {Kg}^{- 1}^{\circ} C^{- 1})$
$∴$ Number of units of electric energy,
$N = \frac{Q}{1 kWh} = \frac{1449 \times 10^{3}}{3.6 \times 10^{6}} (∴ 1 kWh = 3.6 \times 10^{6} J)$
$N = 0.40$ unit
Cost of electric energy in $R_{s} = C \times N$
$= 25 \times 0.40$
$= 10 Paisa$
Therefore, the cost of heating is 10 Paisa.

AP EAMCET (22.04.2019) Shift-II

Thermal Properties of Matter

146651 The specific heat capacities of three liquids $A$ ,
$B$ and $C$ are in the ratio, $1 : 2 : 3$ and the masses of the liquids are in the ratio $1 : 1 : 1$ . The temperatures of the liquids $A, B$ and $C$ are $15^{\circ} C, 30^{\circ} C$ and $45^{\circ} C$ , respectively. Then matched the resultant temperature of the mixture given in List-II with the corresponding mixture given in List-I.
| List-I | | List-II | |
| :--- | :--- | :--- | :--- |
| A. | Mixture of liquids A and B | i. | $25^{\circ} C$ |
| B. | Mixture of liquids B and C | ii. | $35^{\circ} C$ |
| C. | Mixture of liquids C and A | iii. | ${37.5}^{\circ} C$ |
| D. | Mixture of liquids A,B and C | iv. | $39^{\circ} C$ |

1 A - (i),B - (ii),C - (iii),D -(iv)

2 A - (ii),B - (i) ,C - (iv),D - (iii)

3 A - (i),B - (iv),C - (iii),D - (ii)

4 A - (iv),B - (i),C - (iii),D - (ii)

Explanation:

C Let the resultant temperature of mixture A and $B$ is $T_{1}$
then, from principle of colorimetry,
heat lost by $B =$ heat gained by $A$
$(∴$ temperature $B = 30^{\circ} C$ and temp. $A = 15^{\circ} C)$
So, $B$ loses heat and $A$ gains heat
$M_{B} C_{B} (30 - T_{1}) = M_{A} C_{A} (T_{1} - 15)$
So, $m \times 2 \times (30 - T_{1}) = m \times 1 \times (T_{1} - 15)$
$2 (30 - T_{1}) = 1 (T_{1} - 15)$
$T_{1} = 25^{\circ} C$
Here,
$\begin{array}{ll} m_{1} = m_{2} = m_{3} = 1 \\ and, c_{1} : c_{2} : c_{3} = 1 : 2 : 3 \end{array}$
For (B)
Let resultant temperature of mixture $B$ and $C$ is $T_{2}$
Then, Heat lost by $C =$ Heat gained by $B$
$3 (45 - T_{2}) = 2 (T_{2} - 30)$
$T_{2} = 39^{\circ} C$
For (C)
Let resultant temperature of mixture $C$ and $A$ is $T_{3}$ then, Heat lost by $C =$ heat gained by $A$
Similarly, $3 (45 - T_{3}) = (T_{3} - 15) \Rightarrow T_{3} = {37.5}^{\circ} C$
Hence, $A \to$ (i), B $\to$ (iv), C $\to$ (iii), D $\to$ (ii)

AP EAMCET (22.04.2019) Shift-II

Thermal Properties of Matter

146648 If $60 %$ of the kinetic energy of water falling from $210 m$ high water fall is converted into heat. The raise in temperature of water at the bottom of the falls is nearly (specific heat of water $= 4.2 \times 10^{3} {Jkg}^{- 1} K^{- 1}$ )

1

0.6^{\circ} C

2

{0.3}^{\circ} C

3

1.2 K

4

2.4 K

Explanation:

B
Exp:
Given, $h = 210 m$
Specific heat of waters $= 4.2 \times 10^{3} J {kg}^{- 1} K^{- 1}$
As we know that,
When water falls on the surface of earth, then its potential energy is converted into kinetic energy.
$∴$ Kinetic energy $=$ Potential energy
$K . E = mgh$
According to question,
$mc Δ T = \frac{60}{100} \times mgh$
$Δ T = \frac{0.6 gh}{s} = \frac{0.6 \times 10 \times 210}{4.2 \times 10^{3}}$
$Δ T = {0.3}^{\circ} C$

AP EAMCET (18.09.2020) Shift-I

Thermal Properties of Matter

146649 The specific heat capacities of an ideal gas at the constant pressure and at constant volume are $620 {Jkg}^{- 1} k^{- 1}$ and $420 {Jkg}^{- 1} K^{- 1}$ Respectively. The density of the gas at STP is approximately,

1

2.88 {kgm}^{- 3}

2

4.86 {kgm}^{- 3}

3

3.88 {kgm}^{- 3}

4

1.86 {kgm}^{- 3}

Explanation:

D Given,
Specific heat capacity of a gas at constant pressure, $c_{p} =$ $620 {Jkg}^{- 1} K^{- 1}$
Specific heat capacity of gas at constant volume, $c_{V} =$ $420 {Jkg}^{- 1} K^{- 1}$
$∴$ Molar specific heat of gas at constant pressure and constant volume are given by
$∴ c_{p} = {mc}_{p} = m (620)$
$c_{v} = {mc}_{p} = m (420)$
As we know that,
$∴ c_{P} - c_{V} = R$ (Here, $R =$ universal gas constant)
$m (620 - 420) = R$
$m = \frac{R}{200}$
Where, $μ =$ number of molar of gas
$PV = RT (∴$ for 1 mole, $μ = 1)$
$Pm = ρ RT$
( $∴$ density, $ρ = \frac{m}{V}$ )
$10^{5} \times \frac{R}{200} = ρ R (273)$
${STP, P = 1 \times 10^{5} atm, T = 273 K}$
$ρ = 1.85 {kgm}^{- 3}$

AP EAMCET (23.04.2019) Shift-I

Thermal Properties of Matter

146650 Electrical energy costs 25 paisa per kilowatt hour. Assuming that no energy is wasted, the cost of heating $4.6 kg$ of water from $25^{\circ} C$ to the boiling point is

1 25 paisa

2 50 paisa

3 20 paisa

4 10 paisa

Explanation:

D Given,
Electric energy $cost (C) = 25$ paisa/ $KWh$
Mass of water, $m = 4.6 kg$
Initial temperature of water $T_{i} = 25^{\circ} C$
Final temperature of water $T_{f} = 100^{\circ} C$ (Boiling point of water) as heat used by water to boil,
$Q = ms (T_{f} - T_{i})$
So, $Q = 4.6 \times 4.2 \times 10^{3} \times (100 - 25)$
$Q = 1449 \times 10^{3} J (∴ s = 4.2 \times 10^{3} J {Kg}^{- 1}^{\circ} C^{- 1})$
$∴$ Number of units of electric energy,
$N = \frac{Q}{1 kWh} = \frac{1449 \times 10^{3}}{3.6 \times 10^{6}} (∴ 1 kWh = 3.6 \times 10^{6} J)$
$N = 0.40$ unit
Cost of electric energy in $R_{s} = C \times N$
$= 25 \times 0.40$
$= 10 Paisa$
Therefore, the cost of heating is 10 Paisa.

AP EAMCET (22.04.2019) Shift-II

Thermal Properties of Matter

146651 The specific heat capacities of three liquids $A$ ,
$B$ and $C$ are in the ratio, $1 : 2 : 3$ and the masses of the liquids are in the ratio $1 : 1 : 1$ . The temperatures of the liquids $A, B$ and $C$ are $15^{\circ} C, 30^{\circ} C$ and $45^{\circ} C$ , respectively. Then matched the resultant temperature of the mixture given in List-II with the corresponding mixture given in List-I.
| List-I | | List-II | |
| :--- | :--- | :--- | :--- |
| A. | Mixture of liquids A and B | i. | $25^{\circ} C$ |
| B. | Mixture of liquids B and C | ii. | $35^{\circ} C$ |
| C. | Mixture of liquids C and A | iii. | ${37.5}^{\circ} C$ |
| D. | Mixture of liquids A,B and C | iv. | $39^{\circ} C$ |

1 A - (i),B - (ii),C - (iii),D -(iv)

2 A - (ii),B - (i) ,C - (iv),D - (iii)

3 A - (i),B - (iv),C - (iii),D - (ii)

4 A - (iv),B - (i),C - (iii),D - (ii)

Explanation:

C Let the resultant temperature of mixture A and $B$ is $T_{1}$
then, from principle of colorimetry,
heat lost by $B =$ heat gained by $A$
$(∴$ temperature $B = 30^{\circ} C$ and temp. $A = 15^{\circ} C)$
So, $B$ loses heat and $A$ gains heat
$M_{B} C_{B} (30 - T_{1}) = M_{A} C_{A} (T_{1} - 15)$
So, $m \times 2 \times (30 - T_{1}) = m \times 1 \times (T_{1} - 15)$
$2 (30 - T_{1}) = 1 (T_{1} - 15)$
$T_{1} = 25^{\circ} C$
Here,
$\begin{array}{ll} m_{1} = m_{2} = m_{3} = 1 \\ and, c_{1} : c_{2} : c_{3} = 1 : 2 : 3 \end{array}$
For (B)
Let resultant temperature of mixture $B$ and $C$ is $T_{2}$
Then, Heat lost by $C =$ Heat gained by $B$
$3 (45 - T_{2}) = 2 (T_{2} - 30)$
$T_{2} = 39^{\circ} C$
For (C)
Let resultant temperature of mixture $C$ and $A$ is $T_{3}$ then, Heat lost by $C =$ heat gained by $A$
Similarly, $3 (45 - T_{3}) = (T_{3} - 15) \Rightarrow T_{3} = {37.5}^{\circ} C$
Hence, $A \to$ (i), B $\to$ (iv), C $\to$ (iii), D $\to$ (ii)

AP EAMCET (22.04.2019) Shift-II

Thermal Properties of Matter

146648 If $60 %$ of the kinetic energy of water falling from $210 m$ high water fall is converted into heat. The raise in temperature of water at the bottom of the falls is nearly (specific heat of water $= 4.2 \times 10^{3} {Jkg}^{- 1} K^{- 1}$ )

1

0.6^{\circ} C

2

{0.3}^{\circ} C

3

1.2 K

4

2.4 K

Explanation:

B
Exp:
Given, $h = 210 m$
Specific heat of waters $= 4.2 \times 10^{3} J {kg}^{- 1} K^{- 1}$
As we know that,
When water falls on the surface of earth, then its potential energy is converted into kinetic energy.
$∴$ Kinetic energy $=$ Potential energy
$K . E = mgh$
According to question,
$mc Δ T = \frac{60}{100} \times mgh$
$Δ T = \frac{0.6 gh}{s} = \frac{0.6 \times 10 \times 210}{4.2 \times 10^{3}}$
$Δ T = {0.3}^{\circ} C$

AP EAMCET (18.09.2020) Shift-I

Thermal Properties of Matter

146649 The specific heat capacities of an ideal gas at the constant pressure and at constant volume are $620 {Jkg}^{- 1} k^{- 1}$ and $420 {Jkg}^{- 1} K^{- 1}$ Respectively. The density of the gas at STP is approximately,

1

2.88 {kgm}^{- 3}

2

4.86 {kgm}^{- 3}

3

3.88 {kgm}^{- 3}

4

1.86 {kgm}^{- 3}

Explanation:

D Given,
Specific heat capacity of a gas at constant pressure, $c_{p} =$ $620 {Jkg}^{- 1} K^{- 1}$
Specific heat capacity of gas at constant volume, $c_{V} =$ $420 {Jkg}^{- 1} K^{- 1}$
$∴$ Molar specific heat of gas at constant pressure and constant volume are given by
$∴ c_{p} = {mc}_{p} = m (620)$
$c_{v} = {mc}_{p} = m (420)$
As we know that,
$∴ c_{P} - c_{V} = R$ (Here, $R =$ universal gas constant)
$m (620 - 420) = R$
$m = \frac{R}{200}$
Where, $μ =$ number of molar of gas
$PV = RT (∴$ for 1 mole, $μ = 1)$
$Pm = ρ RT$
( $∴$ density, $ρ = \frac{m}{V}$ )
$10^{5} \times \frac{R}{200} = ρ R (273)$
${STP, P = 1 \times 10^{5} atm, T = 273 K}$
$ρ = 1.85 {kgm}^{- 3}$

AP EAMCET (23.04.2019) Shift-I

Thermal Properties of Matter

146650 Electrical energy costs 25 paisa per kilowatt hour. Assuming that no energy is wasted, the cost of heating $4.6 kg$ of water from $25^{\circ} C$ to the boiling point is

1 25 paisa

2 50 paisa

3 20 paisa

4 10 paisa

Explanation:

D Given,
Electric energy $cost (C) = 25$ paisa/ $KWh$
Mass of water, $m = 4.6 kg$
Initial temperature of water $T_{i} = 25^{\circ} C$
Final temperature of water $T_{f} = 100^{\circ} C$ (Boiling point of water) as heat used by water to boil,
$Q = ms (T_{f} - T_{i})$
So, $Q = 4.6 \times 4.2 \times 10^{3} \times (100 - 25)$
$Q = 1449 \times 10^{3} J (∴ s = 4.2 \times 10^{3} J {Kg}^{- 1}^{\circ} C^{- 1})$
$∴$ Number of units of electric energy,
$N = \frac{Q}{1 kWh} = \frac{1449 \times 10^{3}}{3.6 \times 10^{6}} (∴ 1 kWh = 3.6 \times 10^{6} J)$
$N = 0.40$ unit
Cost of electric energy in $R_{s} = C \times N$
$= 25 \times 0.40$
$= 10 Paisa$
Therefore, the cost of heating is 10 Paisa.

AP EAMCET (22.04.2019) Shift-II

Thermal Properties of Matter

146651 The specific heat capacities of three liquids $A$ ,
$B$ and $C$ are in the ratio, $1 : 2 : 3$ and the masses of the liquids are in the ratio $1 : 1 : 1$ . The temperatures of the liquids $A, B$ and $C$ are $15^{\circ} C, 30^{\circ} C$ and $45^{\circ} C$ , respectively. Then matched the resultant temperature of the mixture given in List-II with the corresponding mixture given in List-I.
| List-I | | List-II | |
| :--- | :--- | :--- | :--- |
| A. | Mixture of liquids A and B | i. | $25^{\circ} C$ |
| B. | Mixture of liquids B and C | ii. | $35^{\circ} C$ |
| C. | Mixture of liquids C and A | iii. | ${37.5}^{\circ} C$ |
| D. | Mixture of liquids A,B and C | iv. | $39^{\circ} C$ |

1 A - (i),B - (ii),C - (iii),D -(iv)

2 A - (ii),B - (i) ,C - (iv),D - (iii)

3 A - (i),B - (iv),C - (iii),D - (ii)

4 A - (iv),B - (i),C - (iii),D - (ii)

Explanation:

C Let the resultant temperature of mixture A and $B$ is $T_{1}$
then, from principle of colorimetry,
heat lost by $B =$ heat gained by $A$
$(∴$ temperature $B = 30^{\circ} C$ and temp. $A = 15^{\circ} C)$
So, $B$ loses heat and $A$ gains heat
$M_{B} C_{B} (30 - T_{1}) = M_{A} C_{A} (T_{1} - 15)$
So, $m \times 2 \times (30 - T_{1}) = m \times 1 \times (T_{1} - 15)$
$2 (30 - T_{1}) = 1 (T_{1} - 15)$
$T_{1} = 25^{\circ} C$
Here,
$\begin{array}{ll} m_{1} = m_{2} = m_{3} = 1 \\ and, c_{1} : c_{2} : c_{3} = 1 : 2 : 3 \end{array}$
For (B)
Let resultant temperature of mixture $B$ and $C$ is $T_{2}$
Then, Heat lost by $C =$ Heat gained by $B$
$3 (45 - T_{2}) = 2 (T_{2} - 30)$
$T_{2} = 39^{\circ} C$
For (C)
Let resultant temperature of mixture $C$ and $A$ is $T_{3}$ then, Heat lost by $C =$ heat gained by $A$
Similarly, $3 (45 - T_{3}) = (T_{3} - 15) \Rightarrow T_{3} = {37.5}^{\circ} C$
Hence, $A \to$ (i), B $\to$ (iv), C $\to$ (iii), D $\to$ (ii)

AP EAMCET (22.04.2019) Shift-II

Thermal Properties of Matter

146648 If $60 %$ of the kinetic energy of water falling from $210 m$ high water fall is converted into heat. The raise in temperature of water at the bottom of the falls is nearly (specific heat of water $= 4.2 \times 10^{3} {Jkg}^{- 1} K^{- 1}$ )

1

0.6^{\circ} C

2

{0.3}^{\circ} C

3

1.2 K

4

2.4 K

Explanation:

B
Exp:
Given, $h = 210 m$
Specific heat of waters $= 4.2 \times 10^{3} J {kg}^{- 1} K^{- 1}$
As we know that,
When water falls on the surface of earth, then its potential energy is converted into kinetic energy.
$∴$ Kinetic energy $=$ Potential energy
$K . E = mgh$
According to question,
$mc Δ T = \frac{60}{100} \times mgh$
$Δ T = \frac{0.6 gh}{s} = \frac{0.6 \times 10 \times 210}{4.2 \times 10^{3}}$
$Δ T = {0.3}^{\circ} C$

AP EAMCET (18.09.2020) Shift-I

Thermal Properties of Matter

146649 The specific heat capacities of an ideal gas at the constant pressure and at constant volume are $620 {Jkg}^{- 1} k^{- 1}$ and $420 {Jkg}^{- 1} K^{- 1}$ Respectively. The density of the gas at STP is approximately,

1

2.88 {kgm}^{- 3}

2

4.86 {kgm}^{- 3}

3

3.88 {kgm}^{- 3}

4

1.86 {kgm}^{- 3}

Explanation:

D Given,
Specific heat capacity of a gas at constant pressure, $c_{p} =$ $620 {Jkg}^{- 1} K^{- 1}$
Specific heat capacity of gas at constant volume, $c_{V} =$ $420 {Jkg}^{- 1} K^{- 1}$
$∴$ Molar specific heat of gas at constant pressure and constant volume are given by
$∴ c_{p} = {mc}_{p} = m (620)$
$c_{v} = {mc}_{p} = m (420)$
As we know that,
$∴ c_{P} - c_{V} = R$ (Here, $R =$ universal gas constant)
$m (620 - 420) = R$
$m = \frac{R}{200}$
Where, $μ =$ number of molar of gas
$PV = RT (∴$ for 1 mole, $μ = 1)$
$Pm = ρ RT$
( $∴$ density, $ρ = \frac{m}{V}$ )
$10^{5} \times \frac{R}{200} = ρ R (273)$
${STP, P = 1 \times 10^{5} atm, T = 273 K}$
$ρ = 1.85 {kgm}^{- 3}$

AP EAMCET (23.04.2019) Shift-I

Thermal Properties of Matter

146650 Electrical energy costs 25 paisa per kilowatt hour. Assuming that no energy is wasted, the cost of heating $4.6 kg$ of water from $25^{\circ} C$ to the boiling point is

1 25 paisa

2 50 paisa

3 20 paisa

4 10 paisa

Explanation:

D Given,
Electric energy $cost (C) = 25$ paisa/ $KWh$
Mass of water, $m = 4.6 kg$
Initial temperature of water $T_{i} = 25^{\circ} C$
Final temperature of water $T_{f} = 100^{\circ} C$ (Boiling point of water) as heat used by water to boil,
$Q = ms (T_{f} - T_{i})$
So, $Q = 4.6 \times 4.2 \times 10^{3} \times (100 - 25)$
$Q = 1449 \times 10^{3} J (∴ s = 4.2 \times 10^{3} J {Kg}^{- 1}^{\circ} C^{- 1})$
$∴$ Number of units of electric energy,
$N = \frac{Q}{1 kWh} = \frac{1449 \times 10^{3}}{3.6 \times 10^{6}} (∴ 1 kWh = 3.6 \times 10^{6} J)$
$N = 0.40$ unit
Cost of electric energy in $R_{s} = C \times N$
$= 25 \times 0.40$
$= 10 Paisa$
Therefore, the cost of heating is 10 Paisa.

AP EAMCET (22.04.2019) Shift-II

Thermal Properties of Matter

146651 The specific heat capacities of three liquids $A$ ,
$B$ and $C$ are in the ratio, $1 : 2 : 3$ and the masses of the liquids are in the ratio $1 : 1 : 1$ . The temperatures of the liquids $A, B$ and $C$ are $15^{\circ} C, 30^{\circ} C$ and $45^{\circ} C$ , respectively. Then matched the resultant temperature of the mixture given in List-II with the corresponding mixture given in List-I.
| List-I | | List-II | |
| :--- | :--- | :--- | :--- |
| A. | Mixture of liquids A and B | i. | $25^{\circ} C$ |
| B. | Mixture of liquids B and C | ii. | $35^{\circ} C$ |
| C. | Mixture of liquids C and A | iii. | ${37.5}^{\circ} C$ |
| D. | Mixture of liquids A,B and C | iv. | $39^{\circ} C$ |

1 A - (i),B - (ii),C - (iii),D -(iv)

2 A - (ii),B - (i) ,C - (iv),D - (iii)

3 A - (i),B - (iv),C - (iii),D - (ii)

4 A - (iv),B - (i),C - (iii),D - (ii)

Explanation:

C Let the resultant temperature of mixture A and $B$ is $T_{1}$
then, from principle of colorimetry,
heat lost by $B =$ heat gained by $A$
$(∴$ temperature $B = 30^{\circ} C$ and temp. $A = 15^{\circ} C)$
So, $B$ loses heat and $A$ gains heat
$M_{B} C_{B} (30 - T_{1}) = M_{A} C_{A} (T_{1} - 15)$
So, $m \times 2 \times (30 - T_{1}) = m \times 1 \times (T_{1} - 15)$
$2 (30 - T_{1}) = 1 (T_{1} - 15)$
$T_{1} = 25^{\circ} C$
Here,
$\begin{array}{ll} m_{1} = m_{2} = m_{3} = 1 \\ and, c_{1} : c_{2} : c_{3} = 1 : 2 : 3 \end{array}$
For (B)
Let resultant temperature of mixture $B$ and $C$ is $T_{2}$
Then, Heat lost by $C =$ Heat gained by $B$
$3 (45 - T_{2}) = 2 (T_{2} - 30)$
$T_{2} = 39^{\circ} C$
For (C)
Let resultant temperature of mixture $C$ and $A$ is $T_{3}$ then, Heat lost by $C =$ heat gained by $A$
Similarly, $3 (45 - T_{3}) = (T_{3} - 15) \Rightarrow T_{3} = {37.5}^{\circ} C$
Hence, $A \to$ (i), B $\to$ (iv), C $\to$ (iii), D $\to$ (ii)

AP EAMCET (22.04.2019) Shift-II

Question Bank Series

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