150g of could lemonade (same specific heat as water) at 3.0°c is poured into a g

Question

150g of could lemonade (same specific heat as water) at 3.0°c is poured into a glass whose specific heat is 840 J/(kg*C°). the glass has just been washed in a dishwasher , so its temperature is 75°c. after a few minutes the lemonade and the glass reach an equilibrium temperature of 10°c. assume an insulating container so that no heat is lost to the environment.

a) how much heat (in J) is gained by the lemonade

b) how much heat (in J) is lost by the glass?

c) what is the mass of the glass (in g)?

Explanation / Answer

According to principle of caloriemetry

A) heat gained by lemonade = heat lost by glass

heat gained by lemonade = ml*Sl*dT_l

ml is the mass of lemonade = 0.15kg
Sl is the specific heat of lemonade = 4186 JKg^-1 K^-1
dT_l is the change in temparature = 10-3 = 7 degrees

heat gained by lemonade = 0.15*4186*7 = 4395.3 J

B) heat lost by glass = mg*Sg*dT_g= heat gained by lemonade = 4395.3 J

C) heat lost by glass = mg*Sg*dT_g = 4395.3

mass of glass = mg = (4395.3)/(Sg*dT_g) = 4395.3/(840*65) = 0.0805 kg = 80.5 g

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