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Heat Transfer & Calorimetry Test - 3
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Heat Transfer & Calorimetry Test - 3
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  • Question 1/3
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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

    Solutions

    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 2/3
    1 / -0

    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

    Solutions

    L 1 = 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 3/3
    1 / -0

    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

    Solutions

    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

     

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