Question

the temperature of 600 gram of cold water rises by 15 degree Celsius when 300 gram of water at 50 degree Celsius is added to it .the initial temperature of the cold water is..

Answer 1

Heat gained by cold water = Heat loss by hot water
mSΔT = m’SΔT’
m ΔT = m’ ΔT’
600 g × 15 °C = 300 g × ΔT’
ΔT’ = 30 °C

Change in temperature of hot water is 30°C
30°C = 50°C - T ………[∵ Initial Temperature of water was 50°C]
T = 20°C

Final temperature of hot water = Temperature of mixture = 20°C

Given that,
Rise in temperature of cold water = 15°C

20°C - T’ = 15°C
T’ = 5°C

∴ Initial Temperature of cold water is 5°C

Answer 2

Answer:

let final temp be t

temp of 600 g of water =t-15

temp of 300 g of water =50-t

now

heat gained by cold body =heat lost by hot body

so 600×c×t-(t-15)=300×c×(50-t)

diving we get t=20°c

so initial temp of cold body =t-15=20-15=5°c