# 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 15 ^{o} C when 300 g of hot water at 50^{o} 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) = 50^{0} C

Mass of cold water (m2) = 600 g

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

Final temperature = T^{0} C

Change in temperature of cold water (T – T2) = 15^{0} 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 = 20^{0 }C

Change in temperature = 15^{0} C

from equation [1]: (T – T2) = 15^{0} C

=> 20^{0} C – T2= 15^{0} C

Initial temperature of cold water T2= 20^{0} C – 15^{0} C = 5^{0} C