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Calculate the mass of ice needed to cool 150 g of water contained in a calorimeter of mass 50 g at 32°C such that the final temperature is 5°C.

Specific heat capacity of calorimeter = 0.4 J g-1 °C-1

Specific heat capacity of water = 4.2 Jg-1°C-1

Latent heat capacity of ice = 330 Jg-1

Calorimetry

ICSE 2017

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Answer

Given,

Mass of water m1 = 150 g

Mass of calorimeter m2 = 50 g

Temperature of water t1 = 32°C

Final temperature t = 5°C

Let m be the mass of ice added

Heat given by calorimeter Q1 = m2 c2 x (t1 - t)

= 50 x 0.4 x (32 - 5) = 540 J

Heat given by water Q2 = m1 c1 x (t1 - t)

= 150 x 4.2 x (32 - 5) = 17010 J

Heat taken by ice to melt at 0°C = m L

= m x 330 Jg-1

Heat energy taken by the melted ice to to raise it's temperature from 0°C to 5 °C

= m x 4.2 x (5 - 0) = 21m J

Total heat taken by ice = 330m + 21m = 351 m J

By principle of calorimeter, if there is no loss of heat.

351 m = 540 + 17010

m = 17550351\dfrac{17550}{351} = 50 g

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