A piece of copper of mass 120g is heated in an enclosure to a temperature of 125°C. It is then taken out of the enclosure and held in air for half a minute and dropped carefully into a copper calorimeter of mass 105 g containing 200 g of water at 20°C. The temperature of the water rises to 25°C. Calculate the rate at which heat is being lost from the piece of copper when it is held in air .
[Specific heat capacity of water = usual value]
[ Specific heat capacity of copper = 400 J kg-¹]
Answers
Answer:
Explanation:
the answer of this question is 0.5 degree Celsius at North and South is 0.1 degree Celsius at waist is 0.15 degree celsius is 0.20 degree Celsius and total of the Celsius is - 0.35
Answer:
13J/S
Explanation:
The Final temperature of copper is 25°c
The Initial temperature of copper is 125°c
The Final temperature of water is 25°c
The Initial temperature of water is 20°c
The Final temperature of copper calorimeter is 25°c
The Initial temperature of copper calorimeter is 20°c
Mass of Copper (Mc) = 120g = 0.12kg
Mass of Water (Mw) = 200g = 0.2kg
Mass of Copper calorimeter (Mcc) = 105g = 0.105kg
Specific heat capacity of copper (Cc) = 400 J/kg/c
Specific heat capacity of water (Cw) = 4200 J/kg/c
Time spent in air = 60/2 = 30s
Heat Lost by Copper = McCc(∆c)
=0.12x400(125-25)
=48(100)
=4800J
Heat Lost to air and Heat gained by Copper calorimeter and water = Heat lost by Copper
Ha + McCc∆° + MwCw∆° = 4800
Ha + 0.105x400(25-20) + 0.2x4200(25-20) =4800
Ha + 42(5) +840(5) =4800
Ha + 210 + 4200 = 4800
Ha + 4410 = 4800
Ha = 4800-4410
Ha = 390J
Rate of heat loss = 390/30
= 13J/s