physics ncert solutions class 11th

11.14 Mass of the metal, m = 0.20 kg = 200 g

Initial temperature of the metal,  $T_1$ = 150 °C, Final temperature of the metal,  $T_2$ = 40 °C

The water equivalent mass of the calorimeter, m' = 0.025 kg = 25 g

Volume of water, V = 150 cm3

Mass of water, M at T = 27 °C, = 150 $*1=150g$

Fall in metal temperature,  $?$ T = $T_1-T_2$ = 150 – 40 = 110 °C

Specific heat of water,  $C_w$ = 4.186 J/g/ $°$ K

Let the specific heat of metal = C

Then, heat loss by the metal,  $\theta$ = mC $?$ T ……. (i)

Rise in the water of the calorimeter system $?$ T' = 40 – 27 = 13°C

Heat gained by the water and calorimeter system; $\theta \text{'}$ = (M + m') $C_w?$ T' …… (ii)

Now heat lost by the metal = heat gained by the water and calorimeter system

mC $?$ T = (M + m') $C_w?$ T'

C = $\frac{(\mathrm{M}\mathrm{}+\mathrm{}\mathrm{m}\mathrm{\text{'}})C_w?\mathrm{T}\mathrm{\text{'}}}{m?\mathrm{T}}$ = $\frac{(150+25)*4.186*13}{200*110}$ = 0.433 J/g/ $°$ K

11.12 Power of the drilling machine, P = 10 kW= 10 $*{10}^3$ W

Mass of the Aluminium block, m = 8.0 kg = 8 $*{10}^3$ gm

Time for which the machine is used, t = 2.5 minute = 2.5 $*60$ s = 150 s

Specific heat of Aluminium, c = 0.91 J/gK

Let the rise of temperature in the block after drilling be $\delta$ T

Total energy consumed by the drilling machine= P $*t$ = 10 $*{10}^3*150$ J = 1.5 $*{10}^6$ J

It is given 50% of energy is useful.

So useful energy,  $?Q$ = 50% of Pt= 0.5 $*$ 1.5 $*{10}^6=$ 7.5 $*{10}^5$ J

We know,  $?Q$ = mc $?$ T or T = $\frac{?Q}{mc}$ = $\frac{7.5\mathrm{}*{10}^5}{8\mathrm{}*{10}^3*0.91}$ = 103 $?$

Therefore 2.5 minute drilling will raise the block temperature by 103 $?$

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