Question

a calorimeter of mass 60g contains 180g of eater at 29 C. Find the final temperature of the mixture when 37.2g of ice at -10C is added to it.

Answer 1

Energy transfer from higher temperature body to lower temperature body .
This way , energy lost by water and gained by ice .

Let assume water lose the energy in such a way that temperature will be 0°C
Then, Q₁ = mS∆T , here S is specific heat capacity of water
S = 4200 cal/g/°C , m = (60 + 180) g = 240g
so, Q₁ = 240 × 4200 × (29-0) = 2.9232 × 10⁷ cal

Let assume ice increases the temperature at 0°C and melt
Q₂ = m₂S₂∆T + m₂Lv
Where S₂ is specific heat of ice = 2100 cal/g/°C
Lv is the latent heat of fusion = 80 cal/g

Now, Q₂ = 37.2 × 2100 × (0-(-10)) + 37.2 × 80
= 7.812 × 10⁵ + 2976
= 7.84176 × 10⁵ cal

Here you can see Q₁ > Q₂
It means water doesn't decrease the temperature to 0°C
Let the final temperature is T
Then, m₁S₁∆T =( m₂S₂∆T + m₂Lv) + m₂S₁∆T
240 × 4200 × (29 - T) = 7.84176 × 10⁵ + 37.2 × 4200 × T
2.9232 × 10⁷ - 1008000T = 7.84176 × 10⁵ + 156240T
2.8447824 × 10⁷ = 1.164240 × 10⁶T
T = 24.43 °C

Hence, final temperature of mixture is 24.43°C