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) = 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 = 20⁰ C

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