Question

the temperature of 600g of cold water rose by 15 degree celcius when 300g of hot water at 50 degree celcius was added to it. What was the initial temperature of the cold water ?

Answer 1

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Answer 2

Given,

Mass of cold water = 600g

Mass of hot water = 300g

Temperature of hot water = 50°C

Temperature rise of cold water = 15°C

To find,

The initial temperature of the cold water.

Solution,

We can simply solve this mathematical problem by using the following mathematical process.

We know that,

Q = mc∆T

Let, the final temperature = T°C

∆T of hot water = (50-T)°C

And, value of ∆T of cold water is already given, which is 15°C

Now, in this case,

Heat gained by cold water = Heat lost by hot water

Now, by putting the given values on both LHS and RHS according to the formula of Q = mc∆T , we get that,

600 × c × 15 = 300 × c × (50-T)

9000 = 15000-300T [Value of "c" or specific heat capacity of water is same in LHS and RHS, hence they are eliminated in this step from both the side.]

9000+300T = 15000

300T = 15000-9000

300T = 6000

T = 20

So, (Initial temperature of cold water + Change in temperature) = Final temperature

Initial temperature = (Final temperature - Change in temperature) = (20-15) = 5°C

Hence, the initial temperature of the cold water was 5°C