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Physics 5 Marks Questions

Calorimetry · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 3 questions.

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Question 15 Marks
A piece of iron of mass 100g is kept inside a furnace for a long time and then put in a calorimeter of water equivalent 10g containing 240g of water at 20°C. The mixture attains and equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470J kg-1°C-1.
Answer
Given,

Mass of iron = 100g

Water equivalent of calorimeter = 10g

Mass of water = 240gm

Let the temperature of surface be $\theta^\circ\text{C}.$

Specific heat capacity of iron = 470J kg-1°C-1

Total heat gained = Total heat lost

$\Rightarrow\frac{100}{1000}\times470\times(\theta-60^\circ)$

$=\frac{(240+10)}{1000}\times4200\times(60-20)$

$\Rightarrow47\theta-47\times60=25\times42\times40$

$\Rightarrow\theta=\frac{42000+2820}{47}=\frac{44820}{47}$

$\Rightarrow953.61^\circ\text{C}$

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Question 25 Marks
The temperatures of equal masses of three different liquids A, B and C are 12°C, 19°C and 28°C respectively. The temperature when A and B are mixed is 16°C, and when B and C are mixed, it is 23°C. What will be the temperature when A and C are mixed?
Answer
Given,

Temperature of A = 12°C

Temperature of B = 19°C

Temperature of C = 28°C

Temperature of mixture of A and B = 16°C

Temperature of mixture of B and C = 23°C

Let the mass of the mixtures be M and the specific heat capacities of the liquids A, B and C be CA, CB, and CC, respectively.

According to the principle of calorimetry, when A and B are mixed, we get

Heat gained by Liquid A = Heat lost by liquid B

⇒ MCA(16 - 12) = MCB(19 - 16)

⇒ 4MCA = 3MCB

$\Rightarrow\text{MC}_\text{A}=\Big(\frac{3}{4}\Big)\text{MC}_\text{B}\ \dots(1)$

When B and C are mixed,

Heat gained by liquid B = Heat lost by liquid C

⇒ MCB(23 - 19) = MCC(28 - 23)

⇒ 4MCB = 5MCC

$\Rightarrow\text{MC}_\text{C}=\Big(\frac{4}{5}\Big)\text{MC}_\text{B}\ \dots(2)$

When A and C are mixed,

Let the temperature of the mixture be T. Then,

Heat gained by liquid A = Heat lost by liquid C

⇒ MCA (T - 12) = MCC (28 - T)

Using the values of MCA and MCC, we get     

From eqs. (1) and (2),

$\Rightarrow\Big(\frac{3}{4}\Big)\text{MC}_\text{B}(\text{T}-12)=\Big(\frac{4}{5}\Big)\text{MC}_\text{B}(28-\text{T})$

$\Rightarrow\Big(\frac{3}{4}\Big)(\text{T}-12)=\Big(\frac{4}{5}\Big)(28-\text{T})$

$\Rightarrow(3\times5)(\text{T}-12)=(4\times4)(28-\text{T})$

$\Rightarrow15\text{T}-180=448-16\text{T}$

$\Rightarrow31\text{T}=628$

$\Rightarrow\text{T}=\frac{628}{31}=20.253^\circ\text{C}$

$\Rightarrow\text{T}=20.3^\circ\text{C}$

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Question 35 Marks
An aluminium vessel of mass 0.5kg contains 0.2kg of water at 20°C. A block of iron of mass 0.2kg at 100°C is gently put into the water. Find the equilibrium temperature of the mixture. Specific heat capacities of aluminium, iron and water are 910Jkg-1-K-1, 470Jkg-1-K-1 and 4200Jkg-1-K-1 respectively.
Answer
Given,

Mass of aluminium = 0.5kg

Mass of water = 0.2kg

Mass of iron = 0.2kg

Specific heat of aluminium = 910Jkg-1-K-1

Specific heat of iron = 470Jkg-1 K-1

Specific heat of water = 4200J kg-1 K-1

Let the equilibrium temperature of the mixture be T.

Temperature of aluminium and water = 20°C = 273 + 20 = 293K

Temperature of iron = 100°C = 273 + 100 = 373K

Heat lost by iron, H1 = 0.2 × 470 × (373 - T)

Heat gained by water = 0.2 × 4200 × (T - 293)

Heat gained by iron = 0.5 × 910 × (T - 293)

Total heat gained by water and iron, H2 = 0.5 × 910 (T - 293) + 0.2 × 4200 × (T - 293)

 H2 = (T - 293) [0.5 × 910 + 0.2 × 4200]

We know,

Heat gain = Heat lost

⇒ (T - 293)[0.5 × 910 + 0.2 × 4200] = 0.2 × 470 × (373 - T)

⇒ (T - 293)(455 + 840) = 94(373 - T)

$\Rightarrow (\text{T} - 293)\frac{1295}{94}=(373−\text{T}) $

⇒ (T - 293) × 14 = (373 - T)

⇒ 14T - 293 × 14 = 373 - T

⇒ 15T = 373 + 4102 = 4475

$\Rightarrow\text{T}=\frac{4475}{15}=298.33\text{K}\approx298\text{K}$

$\therefore$ T = (298 - 273)°C = 25°C

$\therefore$ Final temperature = 25°C

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5 Marks Questions - Physics STD 11 Science Questions - Vidyadip