Solve the following problems:
A calorimeter has mass 100 g and specific heat 0.1 kcal/ kg°C. It contains 250 gm of liquid at 30°C having a specific heat of 0.4 kcal/kg°C. If we drop a piece of ice of mass 10g at 0°C, What will be the temperature of the mixture?
We will use the following:
The specific heat capacity of water = 4.2 × 103 J
Let the temperature of the mixture be “T”
Then,
As per the heat equation, we know,
Q = mcΔT
Then, heat liberated by water, Q1 will be given as :
Q1 = 250×10-3 kg 4.2 × 103 J/kg0C × (300c – T)
⇒ Q1 = 31,500 J – 1050T
Now, the heat absorbed by ice will be given as, Q2 = mLvap
Q2 = 10g × 80 cal/g
Or Q2 = 800 cal
(∵ 1 cal = 4.2 J)
⇒ Q2 = 3360 J
Now, the heat absorbed by water to change its temperature from 00 C to t0C is given as: mice × c× ΔT
⇒ Q3 = 10×10-3 kg × 4.2× 103J× T
⇒ Q3 = 42 t
Now, total heat absorbed by water will be = Total heat absorbed by water = Q2+ Q3 = (3360 + 42T) J
Now, according to the principle of calorimetry
Heat given to the system = Heat taken from the system
⇒ 31,500 J – 1050T = 3360 + 42T
⇒ 28, 140 = 1050T + 42T
⇒ 28,140 = 1092 T
⇒ T = 25°C
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