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### Question 17. 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?

Question 17.

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 know that 1g = 7.716 Cal
Now,
100 g = 771.6 cal 771.6/1000 = 0.7716 kCal
250 g = 1929 cal = 1929/1000 = 1.929 kCal
10 g = 77.16 cal = 77.16/1000 = 0.07716 kCal

heat of fusion in ice(Latent) = 80 k Cal / Kg
We know the formula:-

Heat lost in the process = (M(mass)× C(heat capacity) × ΔT(change in temperature)) -------→(i)

As we know, the ice dropped there, that's the heat loss.

Therefore the Change in temperature = (30 - T) ) ×Heat heat gained by ice = heat lost by the liquid & calorimeter

Substituting this values in the equation (i)

⇒ the final temperature of the calorimeter + final temperature of liquid = Heat loss due to the ice at 00c

⇒ (30 - T) × 0.7716 + (0.1+ (30 - T) ) × { 1.926 × 0.4 } = {0.07716 ×80 } + { T× 0.07716 × 1}

⇒ { 2.3148 - 0.07716) × T + 23.112 -0.7704) × T } = { 6.1728 + 0.07716 ) ×T }

⇒ {2.3148 +23.112 - 6.1728} = {0.07716 + 0.7704 + 0.07716 }

⇒ 0.92472 T = 19.254

⇒T=20.80c

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