5 Marks Questions

Question 15 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 C_A, C_B, and C_C, 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
⇒ MC_A(16 - 12) = MC_B(19 - 16)
⇒ 4MC_A = 3MC_B
_{$\Rightarrow\text{MC}_\text{A}=\Big(\frac{3}{4}\Big)\text{MC}_\text{B}\ \dots(1)$}
When B and C are mixed,
Heat gained by liquid B = Heat lost by liquid C
⇒ MC_B(23 - 19) = MC_C(28 - 23)
⇒ 4MC_B = 5MC_C
_{$\Rightarrow\text{MC}_\text{C}=\Big(\frac{4}{5}\Big)\text{MC}_\text{B}\ \dots(2)$}
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
⇒ MC_A (T - 12) = MC_C (28 - T)
Using the values of MC_Aand MC_C, 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 25 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^-¹K^-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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Question 35 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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