Question 1
To find centripetal acceleration, use the formula : centripetal acceleration = v^2/r
so answer would be (3.71)^2/42.85=0.32 (2d.p.)
Question 2
Force =ma
a= (9.98)^2/31.77=3.1350
Force= 3.1350 * 56.63 = 177.54 (2 d.p.)
Answer: It would be 75
Explanation: If your pushing on the floor at a rate of 5 for 15 seconds you would multiply the numbers and get 75
Complete Question:
A 59.1 g sample of iron is put into a calorimeter (see sketch attached) that contains 100.0 g of water. The iron sample starts off at 85.0 °C and the temperature of the water starts off at 23.0 °C. When the temperature of the water stops changing it's 27.6 °C. The pressure remains constant at 1 atm.
Calculate the specific heat capacity of iron according to this experiment. Be sure your answer is rounded to the correct number of significant digits
(Question attached)
Answer:
![c_{iron}=0.568 J/kg.\°C](https://tex.z-dn.net/?f=c_%7Biron%7D%3D0.568%20J%2Fkg.%5C%C2%B0C)
(rounded to 1 decimal place)
Explanation:
A calorimeter is used to measure the heat of chemical or physical reactions. The example given in the question is using the calorimeter to determine the specific heat capacity of iron.
When the system reaches equilibrium the iron and water will be the same temperature,
. The energy lost from the iron will be equal to the energy gained by the water. It is assumed that the only heat exchange is between the iron and water and no exchange with the surroundings.
(Eq 1)
(Eq 2)
Water:
![m_{water}=100.0 g, c_{water}=4.186 J/kg.\°C, T_{initial,water}=23 \°C, T_{e}=27.6 \°C](https://tex.z-dn.net/?f=m_%7Bwater%7D%3D100.0%20g%2C%20c_%7Bwater%7D%3D4.186%20J%2Fkg.%5C%C2%B0C%2C%20T_%7Binitial%2Cwater%7D%3D23%20%5C%C2%B0C%2C%20T_%7Be%7D%3D27.6%20%5C%C2%B0C)
Iron:
![m_{iron}=59.1 g, c_{iron} = ? J/kg.\°C, T_{initial,iron}=85 \°C, T_{e}=27.6 \°C](https://tex.z-dn.net/?f=m_%7Biron%7D%3D59.1%20g%2C%20c_%7Biron%7D%20%3D%20%3F%20J%2Fkg.%5C%C2%B0C%2C%20T_%7Binitial%2Ciron%7D%3D85%20%5C%C2%B0C%2C%20T_%7Be%7D%3D27.6%20%5C%C2%B0C)
Substituting Eq 1 into Eq 2 and details extracted from the question:
![-m_{iron}c_{iron}(T_{iron,e}-T_{initial})=m_{water}c_{water}(T_{water,e}-T_{initial})](https://tex.z-dn.net/?f=-m_%7Biron%7Dc_%7Biron%7D%28T_%7Biron%2Ce%7D-T_%7Binitial%7D%29%3Dm_%7Bwater%7Dc_%7Bwater%7D%28T_%7Bwater%2Ce%7D-T_%7Binitial%7D%29)
![-59.1*c_{iron}(27.6-85)=100.0*4.186(27.6-23)](https://tex.z-dn.net/?f=-59.1%2Ac_%7Biron%7D%2827.6-85%29%3D100.0%2A4.186%2827.6-23%29)
![c_{iron}=0.568 J/kg.\°C](https://tex.z-dn.net/?f=c_%7Biron%7D%3D0.568%20J%2Fkg.%5C%C2%B0C)
![c_{iron}=0.6 J/kg.\°C](https://tex.z-dn.net/?f=c_%7Biron%7D%3D0.6%20J%2Fkg.%5C%C2%B0C)
Answer:
4276.98 years
Explanation:
t = age of the sample in numbers of years
T = half life of the carbon-14 isotope = 5730 yrs
λ = decay constant of carbon-14
decay constant is given as
![\lambda =\frac{0.693}{T}](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cfrac%7B0.693%7D%7BT%7D)
![\lambda =\frac{0.693}{5730}](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cfrac%7B0.693%7D%7B5730%7D)
![\lambda = 0.000121](https://tex.z-dn.net/?f=%5Clambda%20%3D%200.000121%20)
A₀ = activity of Carbon-14 in living plants
A = activity of Carbon-14 after time "t" = (0.596) A₀
Using the equation
![A = A_{o} e^{-\lambda t}](https://tex.z-dn.net/?f=A%20%3D%20A_%7Bo%7D%20e%5E%7B-%5Clambda%20t%7D)
![(0.596) A₀ = A_{o} e^{-0.000121 t}](https://tex.z-dn.net/?f=%20%280.596%29%20A%E2%82%80%20%20%3D%20A_%7Bo%7D%20e%5E%7B-0.000121%20t%7D)
![0.596 = e^{-0.000121 t}](https://tex.z-dn.net/?f=%200.596%20%3D%20e%5E%7B-0.000121%20t%7D)
t = 4276.98 years