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 ?
1) What is the heat gained by the water and calorimeter?
iii) What is the principle of calorimetry ?
iv) Calculate the specific heat of lead shots ?​

Answer 1

Answer:

Specific heat capacity of the lead ball is 0.0305 cal/g-K

Explanation:

Heat lost by the lead is given as

$Q_{lost} = ms\Delta T$

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

$Q_{lost} = 6600s$

Now heat gained by the water

$Q_1 = m_1s_1\Delta T_1$

$Q_1 = 50(1)(34 - 30)$

$Q_1 = 200 Cal$

Heat gained by calorimeter is given as

$Q_2 = m_2s_2\Delta T_2$

$Q_2 = 3(0.1)(34 - 30)$

$Q_2 = 1.2 cal$

Now by energy balance

Heat gain = Heat loss

$6600 s = 200 + 1.2$

$s = 0.0305 cal/g K$

#Learn

Topic : Calorimeter

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