A 10.9-g iron bullet with a speed of 4.00 102 m/s and a temperature of 20.0° C is stopped in a 0.400-kg block of wood, also at 20.0° C.
(a) At first all of the bullet's kinetic energy goes into the internal energy of the bullet. Calculate the temperature increase of the bullet. (b) After a short time the bullet and the block come to the same temperature T. Calculate T, assuming no heat is lost to the environment.
m1 = 10.9•10^-3 kg, v =400 m/s, t(o) = 20oC, m2 = 0.4 kg,.
specific heat of the iron and wood, respectively,
c1 = 444 J/kg, and c2 = 1700 J/kg.
(a) m1•v^2/2 = 10.9•10^-3•(400)^2/ 2 =872 J = QQ/
Q = m1•c1•Δt,
Δt = Q/ m1•c1 = 872/10.8•10^-3•444 = 180oC.
t(new) = 20oC + 180oC =200oC.
(b)
m1•c1•(200-t) = m2•c2•(t-20),
t = (m1•c1•200 + m2•c2•20)/( m1•c1+ m2•c2) =
(10.9•10^-3•444•200 + 0.4•1700•20)/(10.9•10^-3•444+0.4•1700) =
= 21.27oC
thank you...what I originally did was use the wrong specific heat for iron.
(a) To calculate the temperature increase of the bullet, we can use the equation:
ΔQ = mcΔT
Where:
ΔQ = change in heat energy
m = mass of the bullet (10.9 g = 0.0109 kg)
c = specific heat capacity of iron (0.45 kJ/kg°C or 450 J/kg°C)
ΔT = change in temperature
First, we need to calculate the initial kinetic energy of the bullet:
KE = (1/2)mv^2
Where:
m = mass of the bullet (0.0109 kg)
v = velocity of the bullet (4.00 × 10^2 m/s)
KE = (1/2)(0.0109 kg)(4.00 × 10^2 m/s)^2
KE = 879.68 J
Since all of the bullet's kinetic energy goes into internal energy, we can equate the change in heat energy with the initial kinetic energy:
ΔQ = 879.68 J
Now we can calculate the temperature increase of the bullet:
ΔQ = mcΔT
879.68 J = (0.0109 kg)(450 J/kg°C)ΔT
ΔT = (879.68 J) / [(0.0109 kg)(450 J/kg°C)]
ΔT ≈ 181.045°C
Therefore, the temperature increase of the bullet is approximately 181.045°C.
(b) Once the bullet and block reach the same temperature T, they will have the same internal energy. We can calculate T using the equation:
ΔQbullet = ΔQblock
Where:
ΔQbullet = change in heat energy of the bullet
ΔQblock = change in heat energy of the block
The change in heat energy can be calculated as:
ΔQ = mcΔT
For the bullet:
ΔQbullet = (0.0109 kg)(450 J/kg°C)(T - 20.0°C)
For the block:
ΔQblock = (0.400 kg)(4186 J/kg°C)(T - 20.0°C)
Setting the two equations equal to each other:
(0.0109 kg)(450 J/kg°C)(T - 20.0°C) = (0.400 kg)(4186 J/kg°C)(T - 20.0°C)
Simplifying:
(0.0109 kg)(450 J/kg°C)T - (0.0109 kg)(450 J/kg°C)(20.0°C) = (0.400 kg)(4186 J/kg°C)T - (0.400 kg)(4186 J/kg°C)(20.0°C)
Rearranging the equation:
(0.0109 kg)(450 J/kg°C)T - (0.400 kg)(4186 J/kg°C)T = (0.400 kg)(4186 J/kg°C)(20.0°C) - (0.0109 kg)(450 J/kg°C)(20.0°C)
Simplifying and solving for T:
[(0.0109 kg)(450 J/kg°C) - (0.400 kg)(4186 J/kg°C)]T = (0.400 kg)(4186 J/kg°C)(20.0°C) - (0.0109 kg)(450 J/kg°C)(20.0°C)
T = [(0.400 kg)(4186 J/kg°C)(20.0°C) - (0.0109 kg)(450 J/kg°C)(20.0°C)] / [(0.0109 kg)(450 J/kg°C) - (0.400 kg)(4186 J/kg°C)]
Calculating the value of T using the given values:
T ≈ 327.95°C
Therefore, the temperature at which both the bullet and the block come to equilibrium is approximately 327.95°C.
To solve this problem, we need to understand the concepts of conservation of energy and heat transfer.
(a) The initial kinetic energy of the bullet is given by the equation:
KE = (1/2) * m * v^2
where KE is the kinetic energy, m is the mass of the bullet, and v is its velocity.
Plugging in the given values, we can calculate the initial kinetic energy of the bullet:
KE = (1/2) * (10.9 g) * (4.00 x 10^2 m/s)^2
First, let's convert the mass of the bullet to kilograms:
10.9 g = 10.9 x 10^-3 kg
KE = (1/2) * (10.9 x 10^-3 kg) * (4.00 x 10^2 m/s)^2
Simplifying the equation:
KE = 0.5 * (10.9 x 10^-3) * (4.00 x 10^2)^2
Next, we need to calculate how much of this kinetic energy is transferred into the internal energy of the bullet. We can use the equation:
ΔQ = m * c * ΔT
where ΔQ is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Rearranging the equation:
ΔT = ΔQ / (m * c)
Given that ΔQ is equal to the initial kinetic energy of the bullet and c is the specific heat capacity of iron, we can substitute in the values:
ΔT = KE / (m * c)
Substituting the values and solving for ΔT:
ΔT = (0.5 * (10.9 x 10^-3) * (4.00 x 10^2)^2) / ((10.9 x 10^-3) * c)
To find the specific heat capacity of iron, we need to look it up. The specific heat capacity of iron is approximately 450 J/kg·K.
Using this value and solving for ΔT:
ΔT = (0.5 * (10.9 x 10^-3) * (4.00 x 10^2)^2) / ((10.9 x 10^-3) * 450)
After calculating this expression, you will find the temperature increase of the bullet, ΔT.
(b) To calculate the final temperature T when the bullet and the block come to the same temperature, we can use the equation:
m1 * c1 * ΔT1 = m2 * c2 * ΔT2
Where m1 and c1 are the mass and specific heat capacity of the bullet, ΔT1 is the calculated temperature increase from part (a), m2 is the mass of the block, c2 is the specific heat capacity of the block (which you'll need to look up), and ΔT2 is the change in temperature of the block.
Rearranging the equation:
ΔT2 = (m1 * c1 * ΔT1) / (m2 * c2)
Substituting the known values and solving for ΔT2 will give you the final temperature, T.
Note: In both parts of the question, we assumed that there is no heat lost to the environment. In practice, there may be some heat loss, but this assumption helps simplify the problem for calculation purposes.