Question

120g lead shots heated to a temperature of 89°C is mi s heated to a temperature of 89°C is mixed with water at 30°C kept in a calorimeter. The resultant temperature is found to be 34°C calculate the specific heat of lead shots 2 (mass of calorimeter t of calorimeter 0.1 cal/g °C mass of water = 50g, specific heat of water I cal/g °C)

1) What is the heat lost by the lead shots ?

2) What is the heat gained by the water and calorimeter?

3) What is the principle of calorimetry ?

4) Calculate the specific heat of lead shots ?​

Answer 1

Answer:

Heat lost by the lead is given as

Q_{lost} = ms\Delta TQlost=msΔT

Q_{lost} = 120(s)(89 - 34)Qlost=120(s)(89−34)

Q_{lost} = 6600sQlost=6600s

Now heat gained by the water

Q_1 = m_1s_1\Delta T_1Q1=m1s1ΔT1

Q_1 = 50(1)(34 - 30)Q1=50(1)(34−30)

Q_1 = 200 CalQ1=200Cal

Heat gained by calorimeter is given as

Q_2 = m_2s_2\Delta T_2Q2=m2s2ΔT2

Q_2 = 3(0.1)(34 - 30)Q2=3(0.1)(34−30)

Q_2 = 1.2 calQ2=1.2cal

Now by energy balance

Heat gain = Heat loss

6600 s = 200 + 1.26600s=200+1.2

s = 0.0305 cal/g Ks=0.0305cal/gK

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Topic : Calorimeter