Heat is being supplied at a constant rate to a sphere of ice which is melting at the rate of 0.1 g/sec . It melts completely in 100 sec …

Q: Heat is being supplied at a constant rate to a sphere of ice which is melting at the rate of 0.1 g/sec . It melts completely in 100 sec . The rate of rise of temperature thereafter will be (Assume no loss of heat)

(a) 0.8 °C/sec

(b) 5.4 °C/sec

(c) 3.6 °C/sec

(d) None

Click to See Answer :
Ans: (a)

Sol: m/t = 0.1

m = 0.1 t = 0.1 × 100 = 10 gm

Q = mL

$\displaystyle \frac{dQ}{dt} = L\frac{dm}{dt}$

$\displaystyle \frac{dQ}{dt} = 80 \times 0.1 $

= 8 cal/sec

Again , $\displaystyle \frac{dQ}{dt} = m s \frac{d\theta}{dt} $

$\displaystyle \frac{d\theta}{dt} = \frac{\frac{dQ}{dt}}{m s} $

$\displaystyle \frac{d\theta}{dt} = \frac{8}{10 \times 1}$

= 0.8

 

10 gram of ice at 0 °C is kept in a calorimeter of water equivalent 10 gm . How much heat should be supplied ….

Q: 10 gram of ice at 0 °C is kept in a calorimeter of water equivalent 10 gm . How much heat should be supplied to the apparatus to evaporate the water thus formed ? (Neglect loss of heat)

(a) 6200 cal

(b) 7200 cal

(c) 13600 cal

(d) 8200 cal

Click to See Answer :
Ans: (d)

Sol: Q = mcal scal (100 – 0) + mice[Lf + swater(100-0) + Lv]

= 10 × 1 × 100 + 10[80 + 1×100 + 540]

= 8200 cal

 

A block of mass 2.5 kg is heated to temperature of 500 °C and placed on a large ice block . What is the maximum amount of ice …

Q: A block of mass 2.5 kg is heated to temperature of 500 °C and placed on a large ice block . What is the maximum amount of ice that can be melt (approx.) . Specific heat of the body = 0.1 cal/g°C

(a) 1 kg

(b) 1.5 kg

(c) 2 kg

(d) 2.5 kg

Click to See Answer :
Ans: (b)

Sol: Heat lost by Block = Heat gain by Ice

mblock sblock (500 – 0) = mice,meltedLf

$\displaystyle m_{ice , melted} = \frac{2.5 \times 10^3 \times 0.1 \times 500}{80}$

mice,melted = 1.5 kg

 

A thermally insulated vessel contains some water at 0 °C . The vessel is connected to vacuum pump to pump out water vapour ….

Q: A thermally insulated vessel contains some water at 0 °C . The vessel is connected to vacuum pump to pump out water vapour . This results in some water getting frozen . It is given Latent heat of vaporization of water at 0 °C 21 × 105 J/kg & Latent heat of freezing of water 3.36 × 105 J/kg . The maximum % amount of water that will be solidified in this manner will be

(a) 86.2 %

(b) 33.6 %

(c) 21 %

(d) 24.36 %

Click to See Answer :
Ans: (a)
Sol: $\displaystyle \eta L_f = (1-\eta )L_v $

$\displaystyle \eta = \frac{L_v}{L_f + L_v} $

$\displaystyle \eta = \frac{21 \times 10^5}{3.36 \times 10^5 + 21 \times 10^5} $

$\displaystyle \eta = \frac{21 \times 10^5}{24.36 \times 10^5} = 0.862$

= 86.2 %