Heat transfer between two substances that are in contact is called

A train is traveling at a speed of 75km/hr. The brakes are applied to produce a uniform acceleration of -0.5 m/s2.Find how far t

NNADVOKAT [17]

Answer:

your answer would be 21m

Explanation:

7 0

3 years ago

Before collecting a specimen of urine or feces, the nursing assistant asks the nurse or consults the lab for?

bulgar [2K]

Before collecting a specimen of urine or feces, the nursing assistant asks the nurse or consults the lab for which storage and delivery method to use.

There are different tests that need to be performed on a urine or feces specimen. It depends on the patients that which test would be asked for him to be done.

The storage and delivery methods for urine and feces in the case of different tests has to be consulted by the nurse or consultant.

Incase of an emergency test, whose results are immediately required the method of delivery is a fast mode one. For patients that have a mild disease or are not at risk, other reliable methods can be used. Each specimen should be properly labeled for proper checking.

To learn more about urine specimen, click here:

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8 0

2 years ago

A 0.0550-kg ice cube at −30.0°C is placed in 0.400 kg of 35.0°C water in a very well-insulated container. What is the final temp

KatRina [158]

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 ( $Q_{W}$ ) due to the hot water will be equal to the negative of the one ( $Q_{I}$ ) due to the melted ice.

i.e

$Q_{W}$ = - $Q_{I}$ -----------------(a)

Where;

$Q_{I}$ = $m_{I}$ x $c_{I}$ x Δ $T_{I}$ [ $m_{I}$ = mass of ice, $c_{I}$ = specific heat capacity of melted ice which is now water and Δ $T_{I}$ = change in temperature of the melted ice]

and

$Q_{W}$ = $m_{W}$ x $c_{W}$ x Δ $T_{W}$

[ $m_{W}$ = mass of water, $c_{W}$ = specific heat capacity of water and Δ $T_{W}$ = change in temperature of the water]

Substitute the values of $Q_{W}$ and $Q_{I}$ into equation (a) as follows

$m_{W}$ x $c_{W}$ x Δ $T_{W}$ = - $m_{I}$ x $c_{I}$ x Δ $T_{I}$

Note that $c_{W}$ and $c_{I}$ are the same since they are both specific heat capacities of water. Therefore, the equation above becomes;

$m_{W}$ x Δ $T_{W}$ = - $m_{I}$ x Δ $T_{I}$ -----------------------(b)

Now, let's analyse Δ $T_{W}$ and Δ $T_{I}$ . The final temperature ( $T_{F}$ ) of the two kinds of water(melted ice and cooled water) are now the same.

=> Δ $T_{W}$ = change in temperature of water = final temperature of water( $T_{F}$ ) - initial temperature of water( $T_{IW}$ )

Δ $T_{W}$ = $T_{F}$ - $T_{IW}$

Where;

$T_{IW}$ = 22°C [which is the final temperature of water before mixture]

=> Δ $T_{I}$ = change in temperature of melted ice = final temperature of water( $T_{F}$ ) - initial temperature of melted ice ( $T_{II}$ )

Δ $T_{I}$ = $T_{F}$ - $T_{II}$

$T_{II}$ = 0°C (Initial temperature of the melted ice)

Substitute these values into equation (b) as follows;

$m_{W}$ x Δ $T_{W}$ = - $m_{I}$ x Δ $T_{I}$

0.400 x ( $T_{F}$ - $T_{IW}$ ) = -0.0550 x ( $T_{F}$ - $T_{II}$ )

0.400 x ( $T_{F}$ - 22) = -0.0550 x ( $T_{F}$ - 0)

0.400 x ( $T_{F}$ - 22) = -0.0550 x ( $T_{F}$ )

0.400 $T_{F}$ - 8.8 = -0.0550 $T_{F}$

0.400 $T_{F}$ + 0.0550 $T_{F}$ = 8.8

0.455 $T_{F}$ = 8.8

$T_{F}$ = 19.34°C

Therefore, the final temperature of the mixture is 19.34°C

8 0

3 years ago

A 55 kg boy running at 2.0 m/s jumps onto a 2.0 kg skateboard. Calculate the

Airida [17]

Answer:

The velocity of the boy and the skateboard is 24

5 0

3 years ago

Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length t

gladu [14]

The question is incomplete. The complete question is :

Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).

Mass of the ball : 16.3 g

Predicted range : 0.3503 m

Actual range : 1.09 m

Solution :

Given that :

The predicted range is 0.3503 m

Time of the fall is :

$t=\sqrt{\frac{2H}{g}}$