An iron block of the mass 1 kg is suspended on a spring of the spring constant 140 N/m, and merged into a vessel with 8 liters of water. The mass is displaced by 10 cm from its equilibrium position, and released. How much energy in J has been dissipated by the time the mass comes to a rest?
E = J
.7 OK
HELP: All the initial potential energy has been dissipated.
HELP: The potential energy PE for the spring constant k and displacement x is given by the formula
PE = 1/2*k*x2
(b) What is the mass of water in the container in kg ?
m(water) = kg
8 OK
HELP: 1 liter = 0.001 m3
the density of water is 103 kg/m3
HELP: mass = density*volume
(c) Assuming that the water with the block are thermally isolated from their surroundings, by how much will their temperature increase? Express your answer in C. You will need the values of the heat capacity:
cwater = 4186 J/kg*C
ciron = 448 J/kg*C
Delta T = C
HELP: All the energy from part (a) will be converted to heat.
HELP: If we want to increase the temperature of water by Delta T, we have to increase its thermal energy by
Ewater = cwater*mwater*(Delta T)
The same equation holds for iron if the label water is replaced by iron. But Delta T stays the same, because the temperature of the iron and of the water are the same both at the beginning and at the end.
This hint tells you that the energy from part (a) is equal to
Ewater + Eiron.
So, you can substitute for Ewater and Eiron, and get an equation of the form
Epart (a) = something * (Delta T)
FOr the life of me, I can't figure what you need help with.
(a) To calculate the energy dissipated when the mass comes to a rest, we need to find the initial potential energy of the spring. The formula for the potential energy (PE) of a spring is given by:
PE = (1/2) * k * x^2
Where k is the spring constant and x is the displacement from the equilibrium position.
Given:
Mass of the iron block (m) = 1 kg
Spring constant (k) = 140 N/m
Displacement (x) = 10 cm = 0.1 m
Substituting these values into the formula, we get:
PE = (1/2) * 140 * (0.1)^2
= 0.7 J
So, the initial potential energy of the spring is 0.7 J.
(b) To find the mass of water in the container, we can use the formula:
mass = density * volume
Given:
Volume of water (V) = 8 liters = 8 * 0.001 m^3 (since 1 liter = 0.001 m^3)
Density of water (density) = 103 kg/m^3
Substituting these values into the formula, we get:
mass(water) = 103 * (8 * 0.001)
= 0.824 kg
So, the mass of water in the container is 0.824 kg.
(c) To calculate the increase in temperature of the water and iron, we need to use the equation:
E = c * m * ΔT
Where E is the energy, c is the specific heat capacity, m is the mass, and ΔT is the change in temperature.
Given:
Specific heat capacity of water (cwater) = 4186 J/kg*C
Specific heat capacity of iron (ciron) = 448 J/kg*C
Energy dissipated (E) = 0.7 J
Since all the energy from part (a) is converted to heat, we have:
Ewater + Eiron = E
Substituting the values, we get:
cwater * m(water) * ΔT + ciron * m(iron) * ΔT = E
Since the temperature change is the same for both water and iron, we can rewrite the equation as:
(4186 * 0.824 + 448 * 1) * ΔT = 0.7
Simplifying the equation, we get:
(3441.864 + 448) * ΔT = 0.7
3898.864 * ΔT = 0.7
ΔT = 0.7 / 3898.864
Therefore, the temperature increase (ΔT) is approximately 0.00018°C.
To find the amount of energy dissipated by the time the mass comes to rest in part (a), we can use the formula for potential energy of a spring.
The potential energy (PE) for a spring with a spring constant (k) and displacement (x) is given by the formula:
PE = 1/2 * k * (x^2)
In this case, the spring constant is 140 N/m, and the displacement is 10 cm (0.1 m). Plugging these values into the formula, we get:
PE = 1/2 * 140 N/m * (0.1 m)^2
= 1/2 * 140 N/m * 0.01 m^2
= 0.7 J
Therefore, the amount of energy dissipated by the time the mass comes to rest is 0.7 J.
Now let's move on to part (b), which asks for the mass of water in the container. We're given that the volume of water is 8 liters. To convert this volume to cubic meters, we use the conversion factor 1 liter = 0.001 m^3.
So, the volume of water in cubic meters is:
(volume) = (8 liters) * (0.001 m^3/liter)
= 0.008 m^3
Since we know the density of water is 103 kg/m^3, we can find the mass of water using the formula:
mass = density * volume
mass = (103 kg/m^3) * (0.008 m^3)
= 0.824 kg
Therefore, the mass of water in the container is 0.824 kg.
Moving on to part (c), we're asked to find the change in temperature of the water and the iron when all the energy from part (a) is converted to heat. We're given the heat capacities of water (c_water) and iron (c_iron).
The change in temperature (Delta T) for both water and iron will be the same because their initial and final temperatures are the same.
The equation to calculate the energy required to increase the temperature of a substance is:
E = c * m * Delta T
Where E is the energy, c is the heat capacity, m is the mass, and Delta T is the change in temperature.
Since the initial and final temperatures are the same, the energy from part (a) is equal to the sum of the energy for the water and the iron.
So, we can write the equation as:
0.7 J = c_water * m_water * Delta T + c_iron * m_iron * Delta T
Substituting the known values: c_water = 4186 J/kg*C, m_water = 0.824 kg, c_iron = 448 J/kg*C, and m_iron = 1 kg, we can solve for Delta T.
0.7 J = (4186 J/kg*C) * (0.824 kg) * Delta T + (448 J/kg*C) * (1 kg) * Delta T
Simplifying the equation:
0.7 = 3442.664 * Delta T + 448 * Delta T
Combining like terms:
0.7 = 3890.664 * Delta T
Dividing both sides by 3890.664:
Delta T = 0.7 / 3890.664
= 0.0001797 C
Therefore, the temperature increase of both the water and the iron is approximately 0.0001797 degrees Celsius.