Heat Transfer & Calorimetry Test

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Heat Transfer & Calorimetry Test - 3

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Weekly Quiz Competition

Question 1
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A brass ball of mass 100 g is heated to 100°C and then dropped into 200g of turpentine oil in a calorimeter at 15°C. Final temperature is found to be 23°C. Take specific heat of brass as 0.092 cal/g°C and water equivalent of calorimeter as 4 g.

Heat lost by the ball and specific heat of the brass is approximately

A
810 cal , 0.62cal/g°C

B
610 cal , 0.52cal/g°C

C
710 cal , 0.42cal/g°C

D
510 cal , 0.32cal/g°C

Solution

Let ‘c’ be the specific heat of turpentine

Mass of the solid, M = 100 g

Mass of turpentine, m = 200 g

Water equivalent of calorimeter, W = 4 g

Initial temperature of calorimeter, T₁ = 15°C

Temperature of ball, T₂ = 100°C

Final temperature of the liquid, T = 23°C

Specific heat of solid, c₂ = 0.092 cal/g°C

Heat gained by turpentine and calorimeter

m c (T – T₁) + W (T – T₁)

= 200 c (23 – 15) + 4 (23 – 15)

= (200 c + 4) 8

Heat lost by the ball is

M c₂ (T₂ – T) = 100 (0.092) (100 – 23) = 708.4 cal

From principle of calorimetry :

Heat gained = Heat lost

(200 c + 4) 8 = 708.4

1600 c + 32 = 708.4

c = 708.4−32 / 1600 = 0.42 cal/g°C

The correct answer is: 710 cal , 0.42cal/g°C
Question 2
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A brass rod of length 50 cm and diameter 3 mm is joined to a steel rod of the same length and diameter. What is the change in length (in cm) of the combined rod at 250 °C, if the original lengths are at 40 °C ?

Ends of the rods are free to expand.

Co-efficient of linear expansion of brass = 2.0 × 10^–5 K^–1

Co-efficient of linear expansion of Steel = 1.2 × 10^–5 K ^–1

A
745

B
276

C
834

D
346

Solution

L₁ = 50 cm, α₁ = 2.10 × 10^–5 /°C

ΔT = ( 250 – 40 ) = 210 °C, α₂ = 1.2 × 10^–5 /°C

Change in length of each rod,

ΔL₁ = L₁ α₁ ΔT = 50 × 2.10 × 10^–5 × 210 = 0.22 cm

ΔL₂ = L₂ α₂ ΔT = 50 × 1.2 × 10^–5 × 210 = 0. 126 cm

Total change in length

ΔL = ΔL₁ + ΔL₂ = 0.22 + 0. 1 26 = 346 cm

The correct answer is: 346
Question 3
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A lead bullet just melts when stopped by an obstacle. Assuming that 25 % of the heat is absorbed by the obstacle, find the velocity of the bullet (in m/sec) if its initial temperature is 27°C.

Melting point of lead = 327 °C, specific heat of lead = 0.03 calories /gm/°C, latent heat of fusion of lead = 6 calories / gm, J = 4.2 joules /calorie

A
13

B
15

C
10

D
18

Solution

Given that 25% of the heat is absorbed by the obstacle => 75 % of heat is used in melting lead.

Initial Temperature = 27 °C

Melting Point = 300 °C

(0.75) KE = Heat utilized in increasing the temperature + heat utilized to melt lead at 300°C

(0.75) × 1/2 mv² = M c ΔT + M L

(0.75) × 1/2 v² = ( 0.03 × 300 + 6 ) × 4.2

v = 12.96 m/s = 13 m/s