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 1

Answer:

20.83°C

Explanation:

Solution

Let the final temperature attained be T. then,

heat given by calorimeter + heat given by liquid

= heat absorbed by ice

⇒Mc Sc(30−T)+MLSL (30−T)

=Mi[Lfusion+SW(T−0)]

⇒(100)(0.1)(30−T)+(250)(0.4)(30−T)

=10[80+1(T−0)]

⇒(110)(30−T)=10[80+T]

⇒330−11T=80+T

⇒T= 12(330−80)

=20.83° C

Answer 2

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Let the final temperature of the mixture be T.

Amount of heat required in converting 10 g ice to 0°C to water at 0°C =

10×80=800 cal

Total amount of heat required in converting 10 g water to 0°C to water at T°C =

10×1×T=10T

Total heat energy required to convert 10 g ice at 0°C to water at T°C = 800 + 10T

Amount of heat released to raise the temperature of calorimeter at 30°C to T°C

100×0.1×(30−T)=10(30−T)

Amount of heat released to raise the temperature of 250g of eater at 30°C to T°C =

250×0.4×(30−T)=100(30−T)

Total amount of heat released in the process =

110(30−T)

Using the principle of calorimetry, we have

110(30−T)= 800 + 10T

⇒T=20.83°C