Tag: the essence of change

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Questions Related to the essence of change

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

a piece of lead falls from a height of $100m$ on a fixed non-conducting slab which brings it to rest. If the specific heat of lead is $30.6{\rm{ }}cal/kg{\,^ \circ }C,$, the increase in temperature of the slab immediately after collision is 

  1. ${6.72^ \circ }C$
  2. ${7.62^ \circ }C$
  3. ${5.62^ \circ }C$
  4. ${8.72^ \circ }C$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The potential energy lost by the falling lead piece is converted entirely into heat energy in the slab and lead, assuming no heat loss. Using mgh = m * s * Delta T, where m cancels out, we find Delta T = (g * h) / s. Substituting g = 9.8 or 9.81, h = 100m, and specific heat s = 30.6 cal/kg C (converted to joules if needed, or using direct units where 1 cal = 4.184 J), the temperature increase calculates to approximately 7.62 degrees Celsius.

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

The temperature inside and outside a refrigerator are $273\ K$ and $300\ K$ respectively Assuming that the refrigerator cycle is reversible, for every joule of work done, the heat delivered to the surrounding will be nearly

  1. $11\ J$
  2. $22\ J$
  3. $33\ J$
  4. $50\ J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a reversible refrigerator, the coefficient of performance is COP = T_L / (T_H - T_L) = 273 / (300 - 273) = 273 / 27. The heat delivered to the surroundings is Q_H = W + Q_L. Since COP = Q_L / W, Q_L = COP * W = (273 / 27) * 1 = 273 / 27 J. Then Q_H = W + Q_L = 1 + 273 / 27 = 300 / 27 = 11.11 J, which is nearly 11 J.

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

The height of the Niagara falls is $50$ metres, $(1 cal= 4.2 \mathrm\ { J }).$  Assume its mechanical energy can be completely converted into heat energy.

  1. Heat energy gained by each gram of water is $49 \times 10 ^ { 5 }\ \mathrm { cal }$
  2. Rise in temperature of water is $0.166^{ o }\ C/g$
  3. Rise in temperature of water is $0.12^{ o }\ C$
  4. Heal energy gained is $500\ joule/g$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

The units of fore and length are made three times of their earlier values. Earlier the energy of a system was $81\ J$. What will be the energy of the same system in new units?

  1. $243$
  2. $729$
  3. $9$
  4. $None\ of\ the\ above$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Energy has dimensions [M L^2 T^-2]. If force [M L T^-2] and length [L] are tripled, then [M L T^-2] -> 3 and [L] -> 3. This implies [M] -> 3 / 3 = 1 and [T^-2] -> 3 / 3 = 1 (no change). Thus, energy scales by [M] * [L]^2 * [T^-2] = 1 * 3^2 * 1 = 9. New energy = 81 * 9 = 729 J.

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

When $1\ gm$. of water at $100^{\circ}C$ is converted into steam occupies $1671\ c.c.$ The amount of work done in converting water into steam is

  1. $167\ J$
  2. $180\ J$
  3. $184\ J$
  4. $2098\ J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The work done in a phase change where volume increases is given by W = P * Delta V. Assuming atmospheric pressure P = 1.013 * 10^5 N/m^2 and the change in volume Delta V = 1671 c.c. = 1671 * 10^-6 m^3, multiplying these gives W = (1.013 * 10^5) * (1671 * 10^-6) = 169 J, which is approximately 167 J considering standard approximations for atmospheric pressure and conversion.

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

$1$ calorie is the heat required to increase the temperature of $1g$ of water by $1 ^ { \circ } \mathrm { C }$ from:

  1. $13.5 ^ { \circ } \mathrm { C } \text { to } 14.5 ^ { \circ } \mathrm { C } \text { at } 76 \mathrm { mm } \text { of } \mathrm { Hg }$
  2. $14.5 ^ { \circ } \mathrm { C } \text { to } 15.5 ^ { \circ } \mathrm { C } \text { at } 760 \mathrm { mm } \text { of } \mathrm { Hg }$
  3. $13.5 ^ { \circ } \mathrm { C } \text { to } 15.5 ^ { \circ } \mathrm { Cat } 76 \mathrm { mm } \text { of } \mathrm { Hg }$
  4. $15.5 ^ { \circ } \mathrm { C } \text { to } 16.5 ^ { \circ } \mathrm { C } \text { at } 700 \mathrm { mm } \text { of } \mathrm { Hg }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By international convention, one calorie is defined precisely as the amount of heat required to raise the temperature of 1 gram of pure water from 14.5 degrees Celsius to 15.5 degrees Celsius at a standard pressure of 760 mm of Hg.

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

If the amount of heat given to a system is $35\, J$ and the amount of work done on the system is $15\, J$, then the change in internal energy of the system is

  1. $- 50\, J$
  2. $20\, J$
  3. $30\, J$
  4. $50\, J$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given,

$\Delta Q=+35J$
$\Delta W=-15J$
$\Delta U=?$
From law of thermodynamic,
$\Delta Q=\Delta U+\Delta W$
$\Delta U=\Delta Q-\Delta W$
$\Delta U=35-(-15)$
$\Delta U=35+15$
$\Delta U=50J$
The correct option is D. 

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

A geyser heats water flowing at the rate of 3.0 liters per minute from ${ 27 }^{ \circ  }C$ to ${ 77 }^{ \circ  }C$. If the geyser operates on a gas burner, the rate of consumption of the fuel if its heat of combustion is $4.0\times { 10 }^{ 4 }J/g$ per minute is

  1. $15.75g$
  2. $4 g$
  3. $0.3 g$
  4. $0.16 g$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Heat required = mass * specific heat * delta T = 3000 g * 1 cal/g C * 50 C = 150,000 cal = 630,000 J. Fuel consumption rate = Total heat / Heat of combustion = 630,000 J / 40,000 J/g = 15.75 g.

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

A certain quantity of heat energy is given to a diatomic ideal gas which expands at constant pressure. The fraction of the heat energy that is converted into work is 

  1. $\dfrac 2 5$
  2. $\dfrac 2 7$
  3. $\dfrac 1 5$
  4. $\dfrac 5 7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Supplied heat at constant pressure, 

$\Delta Q=nC _P \Delta T$
Change in internal energy,
$\Delta U=nC _V \Delta T$
ratio, $\dfrac{\Delta U}{\Delta Q}=\dfrac{nC _V \Delta T}{nC _P \Delta T}=\dfrac{C _V}{C _P}$. . . . . . . . .(1)
For diatomic ideal gas,
$C _P=\dfrac{7R}{2}\,  ,  C _V=\dfrac{5R}{2}$
From equation (1), we get
$\dfrac{\Delta U}{\Delta Q}=\dfrac{5R/2}{7R/2}=\dfrac{5}{7}$
$(5/7)^{th}$ part of heat supplied is used to increase internal energy.