Question

2. 300 g of ice at z zero degree centigrade is mixed with 500 g of water at 50 degrees centigrade. What will be the final temperature of the mixture? A. 1.25 celsius B. 2.0 Celsius C. neither A nor B

Answer 1

Solution:

Given info: Mass of water (m_w) ​= 500g
Mass of ice, (m_i) ​ = 300g
Specific heat of water (S_w) ​= 1cal g^{-1 0}C^-1
Latent heat of fusion of ice (L_fi) ​= 80 cal g^-1

Let T be the final temperature of the mixture.
Now, Amount of heat lost by water is
= m_w​S_w​(△T) ​= 500×1×(50−T)

and Amount of heat gained by ice is
= m_i​ L_fi + m_i​ S_w ​(△T) = 300×80 + 300×1×(T−0)

According to principle of calorimetry:
Heat lost = Heat gained
or 500×1×(50−T) = 300×80 + 300×1×(T−0)
or 250−5T = 240 + 3T
or 8T = 10

or T = 1.25 ⁰C

Hence, final temperature will be T = 1.25 ⁰C (option (A))