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PHXI11:THERMAL PROPERTIES OF MATTER

366307 A liquid of mass $m$ and specific heat $c$ is heated to a temperature $2 T$ . Another liquid of mass $m / 2$ and specific heat $2 c$ is heated to temperature $T$ . If these two liquids are mixed, the resulting temperature of the mixture is:

1

(2 / 3) T

2

(8 / 5) T

3

(3 / 5) T

4

(3 / 2) T

Explanation:

Temperature of mixture is given by $T = \frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2}}{m_{1} c_{1} + m_{2} c_{2}} = \frac{m \cdot c \cdot 2 T + \frac{m}{2} \cdot 2 c \cdot T}{m \cdot c \cdot + \frac{m}{2} \cdot 2 c} = \frac{3}{2} T$

PHXI11:THERMAL PROPERTIES OF MATTER

366308 $0.1 m^{3}$ of water at $80^{\circ} C$ is mixed with $0.3 m^{3}$ of water at $60^{\circ} C$ . The final temperature of the mixture is

1

65^{\circ} C

2

70^{\circ} C

3

60^{\circ} C

4

75^{\circ} C

Explanation:

Density of water $= 10^{3} k g / m^{3}$
Let the final temperature of the mixture be $T$ .
Heat lost by water at $80^{\circ} C$
$= 0.1 \times 10^{3} \times c \times (80 - T)$
Heat gained by water at $60^{\circ} C$
$= 0.3 \times 10^{3} \times c \times (T - 60)$
Heat lost $=$ Heat gain
$0.1 \times 10^{3} \times c \times (80 - T)$
$= 0.3 \times 10^{3} \times c \times (T - 60)$ .
$\Rightarrow 1 \times (80 - T) = 3 \times (T - 60)$
$\Rightarrow 4 T = 260 \Rightarrow T = 65^{\circ} C .$

KCET - 2009

PHXI11:THERMAL PROPERTIES OF MATTER

366309 A $10 k W$ drilling machine is used to drill a bore in a small aluminium block of mass $8.0 k g$ . How much is the rise in temperature of the block in 2.5 minutes, assuming $50 %$ of power is used up in heating the machine itself or lost to the surroundings? Specific heat of aluminium $= 0.91 J / g^{\circ} C$ .

1

123^{\circ} C

2

83^{\circ} C

3

103^{\circ} C

4

150^{\circ} C

Explanation:

We are given that power of the machine,
$P = 10 k W = 10^{4} W$
Time for which the machine is used,
$t = 2.5 min = 150 s$
Mass of the aluminium block,
$m = 8 k g = 8 \times 10^{3} g$
Specific heat of aluminium, $c = 0.91 J / g^{\circ} C$
Since $50 %$ of the energy is used up in heating
Energy absorbed by the block, $Q = 0.5 P t$
$= 0.5 \times 10^{4} \times 150 J = 7.5 \times 10^{5} J$
Let $Δ T$ be the rise in temperature of the block.
As $Q = m c Δ T$
$Δ T = \frac{Q}{m c} = \frac{7.5 \times 10^{5} (J)}{8 \times 10^{3} (g) \times 0.91 (J / g^{\circ} C)} = 103^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366310 Four spheres $A, B, C$ and $D$ have their radii in arithmetic progression and the specific heat capacities of their substances are geometric progression. If the ratios of heat capacities of $D$ and $B$ to that of $C$ and $A$ are as $8 : 27$ the ratio of masses of $B$ and $A$ is (Assume same density for all spheres)

1

4 : 1

2

1 : 8

3

1 : 2

4

8 : 1

Explanation:

Let the radii of the spheres be $R$ , $R + a, R + 2 a$ and $R + 3 a$ where $a$ is a constant and the specific heat capacities be $s, s r$ , $s r^{2}$ and $s r^{3}$ where $r$ is another constant.
Given that $(\frac{heat capacity of D}{heat capacity of B}) : (\frac{heat capacity of C}{heat capacity of A}) = 8 : 27$ Heat capacity $=$ mass $\times$ specific heat
$\begin{aligned} [\frac{(R + 3 a)^{3} s r^{3}}{(R + a)^{3} s r}] : [\frac{(R + 2 a)^{3} s r^{2}}{R^{3} s}] = 8 : 27 \\ \Rightarrow & R = a \\ \frac{m_{2}}{m_{1}} = \frac{(4 / 3) π (R + R)^{3} ρ}{(4 / 3) π (R)^{3} ρ} = \frac{8}{1} \end{aligned}$

PHXI11:THERMAL PROPERTIES OF MATTER

366307 A liquid of mass $m$ and specific heat $c$ is heated to a temperature $2 T$ . Another liquid of mass $m / 2$ and specific heat $2 c$ is heated to temperature $T$ . If these two liquids are mixed, the resulting temperature of the mixture is:

1

(2 / 3) T

2

(8 / 5) T

3

(3 / 5) T

4

(3 / 2) T

Explanation:

Temperature of mixture is given by $T = \frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2}}{m_{1} c_{1} + m_{2} c_{2}} = \frac{m \cdot c \cdot 2 T + \frac{m}{2} \cdot 2 c \cdot T}{m \cdot c \cdot + \frac{m}{2} \cdot 2 c} = \frac{3}{2} T$

PHXI11:THERMAL PROPERTIES OF MATTER

366308 $0.1 m^{3}$ of water at $80^{\circ} C$ is mixed with $0.3 m^{3}$ of water at $60^{\circ} C$ . The final temperature of the mixture is

1

65^{\circ} C

2

70^{\circ} C

3

60^{\circ} C

4

75^{\circ} C

Explanation:

Density of water $= 10^{3} k g / m^{3}$
Let the final temperature of the mixture be $T$ .
Heat lost by water at $80^{\circ} C$
$= 0.1 \times 10^{3} \times c \times (80 - T)$
Heat gained by water at $60^{\circ} C$
$= 0.3 \times 10^{3} \times c \times (T - 60)$
Heat lost $=$ Heat gain
$0.1 \times 10^{3} \times c \times (80 - T)$
$= 0.3 \times 10^{3} \times c \times (T - 60)$ .
$\Rightarrow 1 \times (80 - T) = 3 \times (T - 60)$
$\Rightarrow 4 T = 260 \Rightarrow T = 65^{\circ} C .$

KCET - 2009

PHXI11:THERMAL PROPERTIES OF MATTER

366309 A $10 k W$ drilling machine is used to drill a bore in a small aluminium block of mass $8.0 k g$ . How much is the rise in temperature of the block in 2.5 minutes, assuming $50 %$ of power is used up in heating the machine itself or lost to the surroundings? Specific heat of aluminium $= 0.91 J / g^{\circ} C$ .

1

123^{\circ} C

2

83^{\circ} C

3

103^{\circ} C

4

150^{\circ} C

Explanation:

We are given that power of the machine,
$P = 10 k W = 10^{4} W$
Time for which the machine is used,
$t = 2.5 min = 150 s$
Mass of the aluminium block,
$m = 8 k g = 8 \times 10^{3} g$
Specific heat of aluminium, $c = 0.91 J / g^{\circ} C$
Since $50 %$ of the energy is used up in heating
Energy absorbed by the block, $Q = 0.5 P t$
$= 0.5 \times 10^{4} \times 150 J = 7.5 \times 10^{5} J$
Let $Δ T$ be the rise in temperature of the block.
As $Q = m c Δ T$
$Δ T = \frac{Q}{m c} = \frac{7.5 \times 10^{5} (J)}{8 \times 10^{3} (g) \times 0.91 (J / g^{\circ} C)} = 103^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366310 Four spheres $A, B, C$ and $D$ have their radii in arithmetic progression and the specific heat capacities of their substances are geometric progression. If the ratios of heat capacities of $D$ and $B$ to that of $C$ and $A$ are as $8 : 27$ the ratio of masses of $B$ and $A$ is (Assume same density for all spheres)

1

4 : 1

2

1 : 8

3

1 : 2

4

8 : 1

Explanation:

Let the radii of the spheres be $R$ , $R + a, R + 2 a$ and $R + 3 a$ where $a$ is a constant and the specific heat capacities be $s, s r$ , $s r^{2}$ and $s r^{3}$ where $r$ is another constant.
Given that $(\frac{heat capacity of D}{heat capacity of B}) : (\frac{heat capacity of C}{heat capacity of A}) = 8 : 27$ Heat capacity $=$ mass $\times$ specific heat
$\begin{aligned} [\frac{(R + 3 a)^{3} s r^{3}}{(R + a)^{3} s r}] : [\frac{(R + 2 a)^{3} s r^{2}}{R^{3} s}] = 8 : 27 \\ \Rightarrow & R = a \\ \frac{m_{2}}{m_{1}} = \frac{(4 / 3) π (R + R)^{3} ρ}{(4 / 3) π (R)^{3} ρ} = \frac{8}{1} \end{aligned}$

PHXI11:THERMAL PROPERTIES OF MATTER

366307 A liquid of mass $m$ and specific heat $c$ is heated to a temperature $2 T$ . Another liquid of mass $m / 2$ and specific heat $2 c$ is heated to temperature $T$ . If these two liquids are mixed, the resulting temperature of the mixture is:

1

(2 / 3) T

2

(8 / 5) T

3

(3 / 5) T

4

(3 / 2) T

Explanation:

Temperature of mixture is given by $T = \frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2}}{m_{1} c_{1} + m_{2} c_{2}} = \frac{m \cdot c \cdot 2 T + \frac{m}{2} \cdot 2 c \cdot T}{m \cdot c \cdot + \frac{m}{2} \cdot 2 c} = \frac{3}{2} T$

PHXI11:THERMAL PROPERTIES OF MATTER

366308 $0.1 m^{3}$ of water at $80^{\circ} C$ is mixed with $0.3 m^{3}$ of water at $60^{\circ} C$ . The final temperature of the mixture is

1

65^{\circ} C

2

70^{\circ} C

3

60^{\circ} C

4

75^{\circ} C

Explanation:

Density of water $= 10^{3} k g / m^{3}$
Let the final temperature of the mixture be $T$ .
Heat lost by water at $80^{\circ} C$
$= 0.1 \times 10^{3} \times c \times (80 - T)$
Heat gained by water at $60^{\circ} C$
$= 0.3 \times 10^{3} \times c \times (T - 60)$
Heat lost $=$ Heat gain
$0.1 \times 10^{3} \times c \times (80 - T)$
$= 0.3 \times 10^{3} \times c \times (T - 60)$ .
$\Rightarrow 1 \times (80 - T) = 3 \times (T - 60)$
$\Rightarrow 4 T = 260 \Rightarrow T = 65^{\circ} C .$

KCET - 2009

PHXI11:THERMAL PROPERTIES OF MATTER

366309 A $10 k W$ drilling machine is used to drill a bore in a small aluminium block of mass $8.0 k g$ . How much is the rise in temperature of the block in 2.5 minutes, assuming $50 %$ of power is used up in heating the machine itself or lost to the surroundings? Specific heat of aluminium $= 0.91 J / g^{\circ} C$ .

1

123^{\circ} C

2

83^{\circ} C

3

103^{\circ} C

4

150^{\circ} C

Explanation:

We are given that power of the machine,
$P = 10 k W = 10^{4} W$
Time for which the machine is used,
$t = 2.5 min = 150 s$
Mass of the aluminium block,
$m = 8 k g = 8 \times 10^{3} g$
Specific heat of aluminium, $c = 0.91 J / g^{\circ} C$
Since $50 %$ of the energy is used up in heating
Energy absorbed by the block, $Q = 0.5 P t$
$= 0.5 \times 10^{4} \times 150 J = 7.5 \times 10^{5} J$
Let $Δ T$ be the rise in temperature of the block.
As $Q = m c Δ T$
$Δ T = \frac{Q}{m c} = \frac{7.5 \times 10^{5} (J)}{8 \times 10^{3} (g) \times 0.91 (J / g^{\circ} C)} = 103^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366310 Four spheres $A, B, C$ and $D$ have their radii in arithmetic progression and the specific heat capacities of their substances are geometric progression. If the ratios of heat capacities of $D$ and $B$ to that of $C$ and $A$ are as $8 : 27$ the ratio of masses of $B$ and $A$ is (Assume same density for all spheres)

1

4 : 1

2

1 : 8

3

1 : 2

4

8 : 1

Explanation:

Let the radii of the spheres be $R$ , $R + a, R + 2 a$ and $R + 3 a$ where $a$ is a constant and the specific heat capacities be $s, s r$ , $s r^{2}$ and $s r^{3}$ where $r$ is another constant.
Given that $(\frac{heat capacity of D}{heat capacity of B}) : (\frac{heat capacity of C}{heat capacity of A}) = 8 : 27$ Heat capacity $=$ mass $\times$ specific heat
$\begin{aligned} [\frac{(R + 3 a)^{3} s r^{3}}{(R + a)^{3} s r}] : [\frac{(R + 2 a)^{3} s r^{2}}{R^{3} s}] = 8 : 27 \\ \Rightarrow & R = a \\ \frac{m_{2}}{m_{1}} = \frac{(4 / 3) π (R + R)^{3} ρ}{(4 / 3) π (R)^{3} ρ} = \frac{8}{1} \end{aligned}$

PHXI11:THERMAL PROPERTIES OF MATTER

366307 A liquid of mass $m$ and specific heat $c$ is heated to a temperature $2 T$ . Another liquid of mass $m / 2$ and specific heat $2 c$ is heated to temperature $T$ . If these two liquids are mixed, the resulting temperature of the mixture is:

1

(2 / 3) T

2

(8 / 5) T

3

(3 / 5) T

4

(3 / 2) T

Explanation:

Temperature of mixture is given by $T = \frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2}}{m_{1} c_{1} + m_{2} c_{2}} = \frac{m \cdot c \cdot 2 T + \frac{m}{2} \cdot 2 c \cdot T}{m \cdot c \cdot + \frac{m}{2} \cdot 2 c} = \frac{3}{2} T$

PHXI11:THERMAL PROPERTIES OF MATTER

366308 $0.1 m^{3}$ of water at $80^{\circ} C$ is mixed with $0.3 m^{3}$ of water at $60^{\circ} C$ . The final temperature of the mixture is

1

65^{\circ} C

2

70^{\circ} C

3

60^{\circ} C

4

75^{\circ} C

Explanation:

Density of water $= 10^{3} k g / m^{3}$
Let the final temperature of the mixture be $T$ .
Heat lost by water at $80^{\circ} C$
$= 0.1 \times 10^{3} \times c \times (80 - T)$
Heat gained by water at $60^{\circ} C$
$= 0.3 \times 10^{3} \times c \times (T - 60)$
Heat lost $=$ Heat gain
$0.1 \times 10^{3} \times c \times (80 - T)$
$= 0.3 \times 10^{3} \times c \times (T - 60)$ .
$\Rightarrow 1 \times (80 - T) = 3 \times (T - 60)$
$\Rightarrow 4 T = 260 \Rightarrow T = 65^{\circ} C .$

KCET - 2009

PHXI11:THERMAL PROPERTIES OF MATTER

366309 A $10 k W$ drilling machine is used to drill a bore in a small aluminium block of mass $8.0 k g$ . How much is the rise in temperature of the block in 2.5 minutes, assuming $50 %$ of power is used up in heating the machine itself or lost to the surroundings? Specific heat of aluminium $= 0.91 J / g^{\circ} C$ .

1

123^{\circ} C

2

83^{\circ} C

3

103^{\circ} C

4

150^{\circ} C

Explanation:

We are given that power of the machine,
$P = 10 k W = 10^{4} W$
Time for which the machine is used,
$t = 2.5 min = 150 s$
Mass of the aluminium block,
$m = 8 k g = 8 \times 10^{3} g$
Specific heat of aluminium, $c = 0.91 J / g^{\circ} C$
Since $50 %$ of the energy is used up in heating
Energy absorbed by the block, $Q = 0.5 P t$
$= 0.5 \times 10^{4} \times 150 J = 7.5 \times 10^{5} J$
Let $Δ T$ be the rise in temperature of the block.
As $Q = m c Δ T$
$Δ T = \frac{Q}{m c} = \frac{7.5 \times 10^{5} (J)}{8 \times 10^{3} (g) \times 0.91 (J / g^{\circ} C)} = 103^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366310 Four spheres $A, B, C$ and $D$ have their radii in arithmetic progression and the specific heat capacities of their substances are geometric progression. If the ratios of heat capacities of $D$ and $B$ to that of $C$ and $A$ are as $8 : 27$ the ratio of masses of $B$ and $A$ is (Assume same density for all spheres)

1

4 : 1

2

1 : 8

3

1 : 2

4

8 : 1

Explanation:

Let the radii of the spheres be $R$ , $R + a, R + 2 a$ and $R + 3 a$ where $a$ is a constant and the specific heat capacities be $s, s r$ , $s r^{2}$ and $s r^{3}$ where $r$ is another constant.
Given that $(\frac{heat capacity of D}{heat capacity of B}) : (\frac{heat capacity of C}{heat capacity of A}) = 8 : 27$ Heat capacity $=$ mass $\times$ specific heat
$\begin{aligned} [\frac{(R + 3 a)^{3} s r^{3}}{(R + a)^{3} s r}] : [\frac{(R + 2 a)^{3} s r^{2}}{R^{3} s}] = 8 : 27 \\ \Rightarrow & R = a \\ \frac{m_{2}}{m_{1}} = \frac{(4 / 3) π (R + R)^{3} ρ}{(4 / 3) π (R)^{3} ρ} = \frac{8}{1} \end{aligned}$