10.5 g of copper metal is heated to 100°C. The copper is then placed into a cup
ID: 2039614 • Letter: 1
Question
10.5 g of copper metal is heated to 100°C. The copper is then placed into a cup that contains 25 g of tap water, initially at 20°C. The final temperature of the mixture is measured and found to be 23°C. Find the specific heat of copper. 1. An ice cube tray holds 570 g of water at 10.0°C. How much heat must the freezer remove to make ice cubes at -10°C? 2. A aluminum calorimeter contains 1.00 kg water. The temperature of the water is measured to be 23°C A 0.3 kg piece of an unknown metal, which was previously heated to 99°C, is added to the water. The system comes to equilibrium at 25°C. Find the specific heat capacity of the 3. unknown metal by applying the principle of conservation of energyExplanation / Answer
Q1.
heat lost by copper=heat gained by the tap water
==>mass of copper*specific heat of copper*temperature change=mass of water*specific heat of water*temperature change
==>10.5*s*(100-23)=25*4.186*(23-20)
==>s=(25*4.186*(23-20))/(10.5*(100-23))=0.38831 J/(gram.degree celcius)
Q2.
heat to be removed=energy required to be removed to take water from 10 degree to 0 degree + energy to be removed to freeze the water at 0 degree + energy to be removed to take ice at 0 degree to -10 degree
=mass*specific heat of water*temperature change+mass*latent heat of fusion+mass*specific heat of ice*temperature change
=0.57*4186*(10-0)+0.57*333*10^3+0.57*2108*(0-(-10))
=2.257*10^5 J
Q3.let specific heat capacity of unknown metal is s J/(kg.degree celsius)
then heat gained by water=heat lost by unknown metal
==>1*4186*(25-23)=0.3*s*(99-25)
==>s=4186*2/(0.3*74)=377.12 J/(kg.K)
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