Calorimetry MCQ Quiz in मल्याळम - Objective Question with Answer for Calorimetry - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Given:

Mass of water in the calorimeter = 9 g

The temperature of water and calorimeter initially = 30°C

The temperature of ice initially = -20°C

The final temperature of the mixture = 5°C

Water equivalent of the calorimeter = 10 g

Concept:

• Heat gave by the hot body = heat gained by the cold body
• Since the final temperature of the mixture is 5° C, it means all the ice must have melted.​
• The water equivalent of a calorimeter is the amount of water that would have caused the same temperature change as the calorimeter for the given heat.

​Formula:

• Heat change of a body during its temperature change: ΔQ = msΔT, where s is the specific heat capacity
• Heat change of a body during phase change: ΔQ = mL, where L is the latent heat

Solution:

Heat given by calorimeter + water -

⇒ ΔQ_given = 9 × 1 × (30 - 5) + 10 × 1 × (30 - 5)

⇒ ΔQgiven = 475 cal      ....(i) 

• Heat gained by the ice (temperature change (-20° to 0°) + phase change + temperature change (0° to -5°),

⇒ ΔQ_gained = m × 0.5 × (0 - (-20)) + m × 80 + m × 1 + (5 - 0)

⇒ ΔQgained = 95m      ...(ii)

• Equating (i) and (ii) -

⇒ 95m = 475

⇒ m = 5 g 

CONCEPT:

• Wiedemann-Franz law states that the ratio between the electronic contribution of the thermal conductivity (k) and electrical conductivity (σ) of metal is proportional to the temperature (T).
• Mathematically it is written as

\(\Rightarrow \frac{{\bf{k}}}{{\bf{\sigma }}} = {\bf{LT}}\)

Or, \(\frac{{\bf{k}}}{{{\bf{\sigma T}}}} = {\bf{L}}\)

Where L = Lorenz number (proportionality constant) and is equal to 2.44 × 10^-8 W Ω K^-2.

EXPLANATION:

• From the above, it is clear that the Wiedemann-Franz law connects the thermal conductivity and electrical conductivity of a material. Therefore option 1 is correct.

• Wiedemann-Franz law quantifies the idea that the metals which are the best electrical conductors are also the best thermal conductors.

CONCEPT:

• Calorimetry means 'measuring heat'.
• When two bodies (one being solid and other liquid or both being liquid) at different temperatures are mixed, heat will be transferred from the body at a higher temperature to a body at lower temperature till both acquire the same temperature.

EXPLANATION:

• When two bodies at different temperatures are mixed, then the body at higher temperature releases heat while the body at lower temperature absorbs it, so that heat loss is equal to the heat gained. Thus, calorimetry represents the law of conservation of heat energy. Hence, option 1 is correct.
• It means the temperature of the mixture can never be lesser than lower temperatures (as a body cannot be cooled below the temperature of the cooling body) and greater than higher temperature (as a body cannot be heated above the temperature of the heating body). Hence option 2 and 3 is incorrect.

Let final temperature be T.

Then

\(m_1S\Delta T = m_2L\)

\(300 \times 1 \times (25 - T) = 100 \times 75\)

\(25 - T = \dfrac{100 \times 75}{300}\)

\(25 - T = 25\)

\(T = 0^oC\)

After that total energy left = 4.7 × 100

Total mass of water = 400 g

Amount of water again converted into ice

\(m = \dfrac{470}{79.7}\)

\(m = 5.9 g\)

Thus whole mass is converted into water at \(0^oC\), and about 5.9 g water is again converted into ice whose temperature is also \(0^oC\).

After achieving the temperature of \(0^oC\), latent heat of fusion is required firstly for conversion of water into ice then further lowering of temperature is possible. So the final temperature will be \(0^oC\).

When A and B are mixed, the common temp becomes \(16^0C\).

Heat lost by B is equal to the heat gained by A.

Given:

\(m_A=m_B=m_C=m\)

Heat lost by B: \(Q_B=m_Bs_B(19-16)=ms_B(19-16)\)

Heat gained by A: \(Q_A=m_As_A(16-12)=ms_A(16-12)\)

\(\therefore ms_B(19-16)=ms_A(16-12)\)

\(\Rightarrow 3s_B=4s_A\)

\(\therefore 12s_B=16s_A\) .....(1)

When B and C are mixed, the common temp becomes \(23^0C\).

Heat lost by C is equal to the heat gained by B.

Heat lost by C: \(Q_C=m_Cs_C(28-23)=5ms_C\)

Heat gained by B: \(Q_B=m_Bs_B(23-19)=4ms_B\)

\(\Rightarrow 4s_B=5s_C\)

\(\therefore 12s_B=15s_C\) .....(2)

From (1) and (2)

\(16s_A=15s_C\)

\(\therefore s_A=\dfrac{15}{16}s_C\)

When A and C are mixed, let T be the common temp.

Heat lost by C: \(Q_C=m_Cs_C(28-T)=ms_C(28-T)\) .....(3)

Heat gained by A: \(Q_B=m_As_A(T-12)=ms_A(T-12)=m\dfrac {15}{16}s_C(T-12)\) ......(4)

Equating (3) and (4)

\(ms_C(28-T) = m\dfrac {15}{16}s_C(T-12)\)

\(\Rightarrow 16(28-T)=15(T-12)\)

\(\Rightarrow 31T= 16*28 + 15*12 = 448 + 180 =628\)

\(\Rightarrow T=20.25^0 \:C\)

\(Q=\dfrac{kATdt}{dx}\) ....(1)

where A is the area of the layer of ice and -T is the temp of the surrounding air. If dm is the mass of water frozen into ice then,

\(Q=dm \cdot L\)

But, \(dm = A\rho dx\) where \(\rho\) is the density of ice.

Hence, \(Q = A\rho L dx\) ....(2)

Equating (1) and (2) we get,

\(dt= \dfrac{\rho L}{kT} xdx\)

\(\Rightarrow \int_{0}^{t}dt=\dfrac{\rho L}{kT} \int_{x_1}^{x_2}xdx\)

\(\Rightarrow t = \dfrac{\rho L}{kT}(x_2^2-x_1^2)\)

\(t_1= \dfrac{\rho L}{kT}(1^2-0^2)\)

\(t_2= \dfrac{\rho L}{kT}(2^2-1^2)\)

\(\therefore \dfrac{t_2}{t_1}= \dfrac{4-1}{1-0}\)

\(\Rightarrow t_2 = 3t_1=3 \cdot 7=21 \text{ hrs}\)

\(\therefore t_2 > 14 \text{ hrs}\)

Concept:

Heat Transfer and Specific Heat Capacity:

• When two liquids of equal mass are mixed, the heat lost by the hotter liquid equals the heat gained by the cooler liquid.
• The formula for heat transfer is:
  Q = mcΔT,
  where:
  Q = heat transferred (J),
  m = mass of the substance (kg),
  c = specific heat capacity (J/kg°C),
  ΔT = change in temperature (°C).
• For two liquids A and B mixed, the equation becomes:
  m × c_A × (T_final - T_A) = m × c_B × (T_B - T_final)
• By applying similar equations to the other mixtures, we can calculate the ratio of the specific heat capacities of liquids A, B, and C.

Calculation:

Given,

Temperature of liquid A, T_A = 15°C

Temperature of liquid B, T_B = 24°C

Temperature of liquid C, T_C = 30°C

When liquids A and B are mixed, the final temperature, T_AB = 20°C

When liquids B and C are mixed, the final temperature, T_BC = 26°C

Let the specific heat capacities of liquids A, B, and C be c_A, c_B, and c_C, respectively.

For the mixture of A and B:

Using the heat balance equation:
m × c_A × (T_AB - T_A) = m × c_B × (T_B - T_AB)

m × c_A × (20 - 15) = m × c_B × (24 - 20)

c_A × 5 = c_B × 4

c_A / c_B = 4 / 5

For the mixture of B and C:

Using the heat balance equation:
m × c_B × (T_BC - T_B) = m × c_C × (T_C - T_BC)

m × c_B × (26 - 24) = m × c_C × (30 - 26)

c_B × 2 = c_C × 4

c_B / c_C = 4 / 2 = 2

From the two equations:

c_A / c_B = 4 / 5

c_B / c_C = 2

Substitute c_B from the second equation into the first equation:

c_A / (c_C / 2) = 4 / 5

c_A × 2 / c_C = 4 / 5

c_A / c_C = 2 / 5

Now, we have the following ratios:

c_A / c_B = 4 / 5

c_B / c_C = 2

c_A / c_C = 2 / 5

Hence, the ratio of specific heat capacities of liquids A, B, and C is:

c_A : c_B : c_C = 8 : 10 : 5

∴ The ratio of specific heat capacities of the liquids A, B, and C is 8:10:5.
Hence, the correct option is 2) 8:10:5.

Let final temperature be T.

Then

\(m_1S\Delta T = m_2L\)

\(300 \times 1 \times (25 - T) = 100 \times 75\)

\(25 - T = \dfrac{100 \times 75}{300}\)

\(25 - T = 25\)

\(T = 0^oC\)

After that total energy left = 4.7 × 100

Total mass of water = 400 g

Amount of water again converted into ice

\(m = \dfrac{470}{79.7}\)

\(m = 5.9 g\)

Thus whole mass is converted into water at \(0^oC\), and about 5.9 g water is again converted into ice whose temperature is also \(0^oC\).

After achieving the temperature of \(0^oC\), latent heat of fusion is required firstly for conversion of water into ice then further lowering of temperature is possible. So the final temperature will be \(0^oC\).

When A and B are mixed, the common temp becomes \(16^0C\).

Heat lost by B is equal to the heat gained by A.

Given:

\(m_A=m_B=m_C=m\)

Heat lost by B: \(Q_B=m_Bs_B(19-16)=ms_B(19-16)\)

Heat gained by A: \(Q_A=m_As_A(16-12)=ms_A(16-12)\)

\(\therefore ms_B(19-16)=ms_A(16-12)\)

\(\Rightarrow 3s_B=4s_A\)

\(\therefore 12s_B=16s_A\) .....(1)

When B and C are mixed, the common temp becomes \(23^0C\).

Heat lost by C is equal to the heat gained by B.

Heat lost by C: \(Q_C=m_Cs_C(28-23)=5ms_C\)

Heat gained by B: \(Q_B=m_Bs_B(23-19)=4ms_B\)

\(\Rightarrow 4s_B=5s_C\)

\(\therefore 12s_B=15s_C\) .....(2)

From (1) and (2)

\(16s_A=15s_C\)

\(\therefore s_A=\dfrac{15}{16}s_C\)

When A and C are mixed, let T be the common temp.

Heat lost by C: \(Q_C=m_Cs_C(28-T)=ms_C(28-T)\) .....(3)

Heat gained by A: \(Q_B=m_As_A(T-12)=ms_A(T-12)=m\dfrac {15}{16}s_C(T-12)\) ......(4)

Equating (3) and (4)

\(ms_C(28-T) = m\dfrac {15}{16}s_C(T-12)\)

\(\Rightarrow 16(28-T)=15(T-12)\)

\(\Rightarrow 31T= 16*28 + 15*12 = 448 + 180 =628\)

\(\Rightarrow T=20.25^0 \:C\)

\(Q=\dfrac{kATdt}{dx}\) ....(1)

where A is the area of the layer of ice and -T is the temp of the surrounding air. If dm is the mass of water frozen into ice then,

\(Q=dm \cdot L\)

But, \(dm = A\rho dx\) where \(\rho\) is the density of ice.

Hence, \(Q = A\rho L dx\) ....(2)

Equating (1) and (2) we get,

\(dt= \dfrac{\rho L}{kT} xdx\)

\(\Rightarrow \int_{0}^{t}dt=\dfrac{\rho L}{kT} \int_{x_1}^{x_2}xdx\)

\(\Rightarrow t = \dfrac{\rho L}{kT}(x_2^2-x_1^2)\)

\(t_1= \dfrac{\rho L}{kT}(1^2-0^2)\)

\(t_2= \dfrac{\rho L}{kT}(2^2-1^2)\)

\(\therefore \dfrac{t_2}{t_1}= \dfrac{4-1}{1-0}\)

\(\Rightarrow t_2 = 3t_1=3 \cdot 7=21 \text{ hrs}\)

\(\therefore t_2 > 14 \text{ hrs}\)