Answer:
229,098.96 J
Explanation:
mass of water (m) = 456 g = 0.456 kg
initial temperature (T) = 25 degrees
final temperature (t) = - 10 degrees
specific heat of ice = 2090 J/kg
latent heat of fusion =33.5 x 10^(4) J/kg
specific heat of water = 4186 J/kg
for the water to be converted to ice it must undergo three stages:
- the water must cool from 25 degrees to 0 degrees, and the heat removed would be Q = m x specific heat of water x change in temp
Q = 0.456 x 4186 x (25 - (-10)) = 66808.56 J
- the water must freeze at 0 degrees, and the heat removed would be Q = m x specific heat of fusion x change in temp
Q = 0.456 x 33.5 x 10^(4) = 152760 J
- the water must cool further to -10 degrees from 0 degrees, and the heat removed would be Q = m x specific heat of ice x change in temp
Q = 0.456 x 2090 x (0 - (-10)) = 9530.4 J
The quantity of heat removed from all three stages would be added to get the total heat removed.
Q total = 66,808.56 + 152,760 + 9,530.4 = 229,098.96 J
Answer:
17 °C
Explanation:
From specific Heat capacity.
Q = cm(t₂-t₁)................. Equation 1
Where Q = Heat absorb by the metal block, c = specific heat capacity of the metal block, m = mass of the metal block, t₂ = final temperature, t₁ = Initial temperature.
make t₁ the subject of the equation
t₁ = t₂-(Q/cm)............... Equation 2
Given: t₂ = 22 °C, Q = 5000 J, m = 4 kg, c = 250 J/kg.°c
Substitute into equation 2
t₁ = 22-[5000/(4×250)
t₁ = 22-(5000/1000)
t₁ = 22-5
t₁ = 17 °C
The magnetic field strength of a very long current-carrying wire is proportional to the inverse of the distance from the wire. The farther you go from the wire, the weaker the magnetic field becomes.
B ∝ 1/d
B = magnetic field strength, d = distance from wire
Calculate the scaling factor for d required to change B from 25μT to 2.8μT:
2.8μT/25μT = 1/k
k = 8.9
You must go to a distance of 8.9d to observe a magnetic field strength of 2.8μT
Velocity=3.4m/sec
Mass=30kg
so kinetic energy=1/2mv^2
=1/2×30×3.4×3.4
=15×3.4×3.4
=15×11.56
=173.4 kg m per second square
Answer:
d) 1/32 microgram
Explanation:
First half life is the time at which the concentration of the reactant reduced to half.
Second half reaction is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/4.
Third half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/8.
Forth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/16.
Fifth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/32.
The initial mass of the sample = 1 microgram
After 5 half-lives, the mass should reduce to 1/32 of the original.
So the concentration left = 1/32 of 1 microgram = 1/32 microgram
