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The temperature of 600 g of cold water rises by 15 C when 300 g of hot water at 50 C is added to it. What was the initial temperature of the cold water?

Calorimetry numerical Question: 12 [Reference Selina ICSE class 10 book exercise]

The temperature of 600 g of cold water rises by 15o C when 300 g of hot water at 50o C is added to it. What was the initial temperature of the cold water?

Solution:

Mass of hot water (m1) = 300 g

Initial Temperature of hot water (T1) = 500 C

Mass of cold water (m2) = 600 g

Say, the initial temperature of cold water = T2 0 C

Final temperature = T0 C

Change in temperature of cold water (T – T2) = 150 C ……………. [1]

The specific heat capacity of water is c

m1 c (T1 – T) = m2 c (T – T2)

300 (50 – T) = 600 (15)

T = 6000 / 300

Final temperature T = 20C

Change in temperature = 150 C

from equation [1]: (T – T2) = 150 C
=> 200 C – T2= 150 C

Initial temperature of cold water T2= 200 C – 150 C = 50 C

See also  Find the time taken by a 500 W  heater to raise the temperature of 50 kg of material of specific heat capacity 960 J kg-1 K-1, from 18 C to 38 C.
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