JAMB Physics Past Questions & Answers - Page 237

1,181.

A good calorimeter should be of

A.

low specific heat capacity and low heat conductivity

B.

low specific heat capacity and high heat conductivity

C.

high specific heat capacity and low heat conductivity

D.

high specific heat capacity and low heat conductivity

E.

dull surface and low heat conductivity

Correct answer is B

No explanation has been provided for this answer.

1,182.

Two thermo flasks of volume Vx and Vy are filled with liquid water at an initial temperature of 0oC. After sometime the temperature were found to be 9x, 9y, respectively. Given that \(\frac{Vx}{Vy}\) = 2 and \(\frac{\theta x}{\theta y}\) = \(\frac{1}{2}\) the ratio of the heat flow into the flask is

A.

\(\frac{1}{4}\)

B.

\(\frac{1}{2}\)

C.

4

D.

1

E.

2

Correct answer is D

Heat = mc\(\theta\)

Hx = Dx x Vx x Cx x \(\theta\)x

Hy = Dy x Vy x Cy x \(\theta\)y

\(\frac{H_x}{H_y}\) = \(\frac{D_x \times V_x \times C_x \times \theta_x}{D_x \times V_x \times C_x \times \theta_x}\)

= 2 x \(\frac{1}{2}\)

= 1

1,183.

Water shows anomalous behaviour.

A.

below 0oC

B.

between 0oC and 4oC

C.

at exactly 4oC

D.

between 4oC and 100oC

E.

above 100oC

Correct answer is B

No explanation has been provided for this answer.

1,184.

In a gas experiment, the pressure of the gas is plotted against the reciprocal of the volume of the gas at a constant temperature. the unit of the slope of the resulting curve is

A.

force

B.

force/m

C.

work

D.

force/m3

E.

energy/m3

Correct answer is C

slope = pressure \(\div\) \(\frac{1}{volume}\)

= \(\frac{N}{m^3}\) x \(\frac{m^3}{1}\)

= Nm = work

1,185.

A quantity of gas occupies a certain volume when the temperature is -73oC and the pressure is 1.5 atmospheres. If the pressure is increased to 4.5 atmospheres and the volume is halved at the same time, what will be the new temperature of the gas?

A.

573oC

B.

327oC

C.

300oC

D.

110oC

E.

27oC

Correct answer is E

General Gas Law:

\(\frac{P_1 V_1}{T_1}\) = \(\frac{P_2 V_2}{T_2}\)

where T\(_1\) = -73 + 273 = 200k,      T\(_2\) = ?,

V\(_1\) = 1,         V\(_2\) = \(\frac{1}{2V}\),

P\(_1\) = 1.5,      P\(_2\) = 4.5

T\(_2\) = \(\frac{P_2 V_2 T_1}{P_1 V_1}\)

T\(_2\) = \(\frac{4.5 * 1/2 * 200}{1.5 X 1}\)

T\(_2\) → 300k or 27ºC