In an experiment to measure the specific heat capacity of copper, 0.02 kg of water at 70°C is poured into a copper calorimeter (
1 answer:
In an experiment to measure the specific heat capacity of copper, 0.02 kg of water at 70°C is poured into a copper calorimeter (with a stirrer) of mass 0.16 kg at 15°C. After stirring, the final temperature is found to be 45°C. If the specific heat of water is 4,200 J/kg/°C,
WATER:-
Mass :- 0.02 kg at 70°C
Specific Heat Of Water is 4,200 J/kg°C,
COPPER:-
MASS:- mass 0.16 kg at 15°C
Temperatures r according to the part !
what is the quantity of heat released per kg of water per 1°C fall in temperature?
Calculate the heat energy released by water in the experiment in cooling from 70°C to 45°C.
Assuming that the heat released by water is entirely used to raise the temperature of calorimeter from 15°C to 45°C) calculate the specific heat capacity of copper.
It is the amount of the heat released by unit mass of the body per 1°C fall in temperature.
It is the amount of the heat absorbed by the unit mass of the body per 1°C rise in temperature
So quantity of heat released per kg of water per 1°C fall in temperature is equal to the specific heat of the water which is 4,200 J/kg°C,
MASS OF WATER --> 0.02 kg
INITIAL TEMPERATURE--> 70°C
FINAL TEMPERATURE--> 45°C
CHANGE IN TEMPERATURE--> (45-70)°C =(- 25)°C
SPECIFIC HEAT OF WATER--> 4200 J/ kg °C
NEGATIVE INDICATES THAT HEAT IS RELEASED BY THE BODY!
IT MEANS COOLING HAS BEEN TAKEN PLACED.
MASS OF COPPER :- 0.16 kg
INITIAL TEMPERATURE:- 15°C
FINAL TEMPERATURE:- 45°C
CHANGE IN TEMPERATURE--> (45-15)°C = 30°C
AMOUNT OF HEAT RELEASED BY WATER --> 2100 J ( From second part)
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Answer:
The final displacement from the origin is <u>1.6</u> km to the <u>NE</u>
Explanation:
The directions in which the delivery truck travels are;
1) 2.8 km North = 2.8·, in vector form
2) 1.0 km East = 1.0·, in vector form
3) 1.6 km South = -1.6·, in vector form
Therefore, to find the final displacement, Δx, of the delivery truck, we add the individual displacements as follows;
Final displacement, Δd = 2.8· + 1.0·
+(-1.6·
) = 1.2·
+ 1.0·
Final displacement, = 1.0· + 1.2·
Where;
Δx = The displacement in the x-direction = 1.0·
Δy = The displacement in the y-direction = 1.2·
The magnitude of the resultant displacement vector is given as follows
= √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)
The magnitude of the resultant displacement vector ≈ 1.6 km
The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE
Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.