Question

pls tell this numerical
5th wala

Answer 1

Let the nal temperature of the mixture be x
 For 500 g of water, m = 500g , T = (80 - x)°C (since both the substances are water we don't require the specific heat capacity)
 So, Q = 500 x (80 - x) J
For 250 g of water,
 m = 250g , T = (x - 50)°C
So, Q = 250 x (x - 50) J
According to the principle of calorimetry,
 heat gained = heat lost
 = 500 x (80 - x) = 250 x (x - 50)
 After solving, we get,
 x= 70
Therefore the nal temperature of the mixture = 70°C

i hope it will help you 
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