Sal connected four light bulbs in a parallel circuit. If Sal adds another light bulb, what is true about the current in each bul
2 answers:
Since all the bulbs are connected in parallel with each other
so here voltage across each bulb will remains the same
so here we will have
now all bulbs are identical and we have same voltage source across all the bulbs
so we will have current in each bulb given by same equation
now we can say that correct answer will be
C) The current in each bulb remained the same even though another bulb was added.
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Answer:
19.34°C
Explanation:
When the ice cube is placed in the water, heat will be transferred from the hot water to it such that the heat gained (Q₁) by the ice is equal to the heat lost(Q₂) by the hot water and a final equilibrium temperature is reached between the melted ice and the cooling/cooled hot water. i.e
Q₁ = -Q₂ ----------------------(i)
{A} Q₁ is the heat gained by the ice and it is given by the sum of ;
(i) the heat required to raise the temperature of the ice from -30°C to 0°C. This is given by [m₁ x c₁ x ΔT]
<em>Where;</em>
m₁ = mass of ice = 0.0550kg
c₁ = a constant called specific heat capacity of ice = 2108J/kg°C
ΔT₁ = change in the temperature of ice as it melts from -30°C to 0°C = [0 - (-30)]°C = [0 + 30]°C = 30°C
(ii) and the heat required to melt the ice completely - This is called the heat of fusion. This is given by [m₁ x L₁]
Where;
m₁ = mass of ice = 0.0550kg
L₁ = a constant called latent heat of fusion of ice = 334 x 10³J/kg
Therefore,
Q₁ = [m₁ x c₁ x ΔT₁] + [m₁ x L₁] ------------------(ii)
Substitute the values of m₁, c₁, ΔT₁ and L₁ into equation (ii) as follows;
Q₁ = [0.0550 x 2108 x 30] + [0.0550 x 334 x 10³]
Q₁ = [3478.2] + [18370]
Q₁ = 21848.2 J
{B} Q₂ is the heat lost by the hot water and is given by
Q₂ = m₂ x c₂ x ΔT₂ -----------------(iii)
Where;
m₂ = mass of water = 0.400kg
c₂ = a constant called specific heat capacity of water = 4200J/Kg°C
ΔT₂ = change in the temperature of water as it cools from 35°C to the final temperature of the hot water (T) = (T - 35)°C
Substitute these values into equation (iii) as follows;
Q₂ = 0.400 x 4200 x (T - 35)
Q₂ = 1680 x (T-35) J
{C} Now to get the final temperature, substitute the values of Q₁ and Q₂ into equation (i) as follows;
Q₁ = -Q₂
=> 21848.2 = - 1680 x (T-35)
=> 35 - T = 21848.2 / 1680
=> 35 - T = 13
=> T = 35 - 13
=> T = 22
Therefore the final temperature of the hot water is 22°C.
Now let's find the final temperature of the mixture.
The mixture contains hot water at 22°C and melted ice at 0°C
At this temperature, the heat () due to the hot water will be equal to the negative of the one (
) due to the melted ice.
i.e
= -
-----------------(a)
Where;
=
x
x Δ
[
= mass of ice,
= specific heat capacity of melted ice which is now water and Δ
= change in temperature of the melted ice]
and
=
x
x Δ
[ = mass of water,
= specific heat capacity of water and Δ
= change in temperature of the water]
Substitute the values of and
into equation (a) as follows
x
x Δ
= -
x
x Δ
Note that and
are the same since they are both specific heat capacities of water. Therefore, the equation above becomes;
x Δ
= -
x Δ
-----------------------(b)
Now, let's analyse Δ and Δ
. The final temperature (
) of the two kinds of water(melted ice and cooled water) are now the same.
=> Δ = change in temperature of water = final temperature of water(
) - initial temperature of water(
)
Δ =
-
Where;
= 22°C [which is the final temperature of water before mixture]
=> Δ = change in temperature of melted ice = final temperature of water(
) - initial temperature of melted ice (
)
Δ =
-
= 0°C (Initial temperature of the melted ice)
Substitute these values into equation (b) as follows;
x Δ
= -
x Δ
0.400 x ( -
) = -0.0550 x (
-
)
0.400 x ( - 22) = -0.0550 x (
- 0)
0.400 x ( - 22) = -0.0550 x (
)
0.400 - 8.8 = -0.0550
0.400 + 0.0550
= 8.8
0.455 = 8.8
= 19.34°C
Therefore, the final temperature of the mixture is 19.34°C
Answer:
a) 0.040625 m
b) 5.02272 J
Explanation:
k = Spring constant
x = Stretched length
F = Force
a)
Extension of the spring would be 0.040625 m
b) Work done in a spring
The work done by the shopper to stretch this spring a total distance of 8.00 cm is 5.02272 J