Question

A calorimeter has mass 100 g and specific heat 0.1 kcal/ kg °C. It contains 250 gm of liquid at 30 °C having specific heat of 0.4 kcal/kg °C. If we drop a piece of ice of mass 10 g at 0 °C, What will be the temperature of the mixture? Answer : 20.8 °C. Solve the problem

Answer 1

First we have to convert the the grams into calories
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 = (mass × heat capacity × change in temperature)  -------→(i)

As we know the ice dropped there, that's the the heat loss. Therefore the Change in temperature = (30 - T)Heat heat gained by ice = heat lost by liquid & calorimeter 

Substituting this values in the equation (i)

⇒ {(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

$\boxed{T = 20.83^0 C}$  

Hence the temperature will be 20.83 ° C.


Hope it Helps :-)

Answer 2

Answer:

20.8℃

is the correct answer