You have 200 g of coffee at 55C, Coffee has the same specific heat as water. How much ice at -5C do you need to add in order to reduce the coffee’s temperature to 27C?
Let m be the mass of ice needed for the purpose.
This is to be solved by using the principle of mixtures or principle of calorimetry. According to this for two isolated bodies in contact with different temperatures, heast lost by hot body is equal to the heat gained by the cold body in achieving equilibrium temperature.
Here specific heat of ice = 0.5 cal/g 0C.
Latent heat of fusion of ice = 80 cal/g
Specific heat of water = 1 cal/ g 0C.
Hence QLost=Qgained . Ice gains and water looses energy.energy.
(m×0.5×(5)) +(m×80)+(m×1×27) = (200×1×(55-27))
m(109.5) = 5600
m = 51.14 gram of ice is needed.
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