Ask question

Ask question

All categories

2 years ago

14

A container contains 200g of water at initial temperature of 30°C. An iron nail of mass 200g at temperature of 50°C is immersed

in the water. What is the final water temperature? State the assumptions you need to make in your calculations.
[Given the value of specific heat capacity of water is 4200 J kg^-1 °C^-1 and that of iron is
450 J kg^-1 °C^-1]

1 answer:

andriy [413]2 years ago

4 0

Answer:

The final temperature is 31.94°

Explanation:

The mass of the water in the container m₁ = 200 g = 0.2 kg

The initial temperature of the water, T₁₁ = 30°C

The mass of the iron, m₂ = 200 g = 0.2 kg

The temperature of the iron T₂₁= 50°C is immersed in the water,

The specific heat capacity of the water, c₁ = 4200 J/(kg·°C)

The specific heat capacity of the iron, c₂ = 450 J/(kg·°C)

Heat capacity relation is given by the formula;

Heat capacity Q = Mass, m × Specific heat capacity, c × Temperature change, (T₂ - T₁)

Given that energy can neither be created nor destroyed, and with the assumption that all the heat lost by the nail is gained by the water we have;

Heat lost by iron nail = Heat gained by the water

m₁ × c₁ × (T₂ - T₁₁) = m₂ × c₂ × (T₂₁ - T₂)

Where, T₂ is the final temperature

0.2 kg × 4200 J/(kg·°C) × (T₂ - 30) = 0.2 kg × 450 J/(kg·°C) × (50° - T₂)

840·T₂ - 25200 = 4500 - 90·T₂

4500 + 25200 = 840·T₂ + 90·T₂

29700 = 930·T₂

T₂ = 29700/930 = 31.94°.

The final temperature = 31.94°.

You might be interested in

A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha

Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3 s

the formula for the time period of the pendulum is given by

$T = 2\pi \sqrt{\frac{I}{mgd}}$ .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

$I = \frac{mL^{2}}{12}+ md^{2}$

Substituting the values in equation (1)

$3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}$

$9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )$

12d² -26.84 d + 2.25 = 0

$d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}$

$d=\frac{26.84\pm 24.75}{24}$

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0

1 year ago

The velocity component with which a projectile covers certain horizontal distance is maximum at the moment of? a) Hitting the gr

Crazy boy [7]

A) hitting the ground

3 0

2 years ago

The length of a 60 W, 240 Ω light bulb filament is 60 cm Remembering that the current in the filament is proportional to the ele

faust18 [17]

Answer:

Finally current will be

i = 0.35 A

Explanation:

As we know that power of the bulb is given by the formula

$P = \frac{V^2}{R}$

now we have

$P = 60 W$

R = 240 ohm

so we have

$60 = \frac{V^2}{240}$

$V = 120 Volts$

now the current in the bulb is given as

$i = \frac{V}{R}$

$i = \frac{120}{240} = 0.5 A$

now when length of the filament is double

so the resistance of the wire also gets double

so we have

$P = \frac{V^2}{R}$

$60 = \frac{V^2}{480}$

$V = 169.7 volts$

now the current in the bulb is given as

$V = i R$

$169.7 = i(480)$

$i = 0.35 A$

8 0

2 years ago

While ice skating, you unintentionally crash into a person. Your mass is 60 kg, and you are traveling east at 8.0 m/s with respe

kaheart [24]

Answer:

6.18 m/s

Explanation:

Roller skate collision

The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;

x-axis component form (+x east);

$P_{Miy}$ + $p_{Piy}$ + $j_{y}$ = $P_{Mfy}$ + $P_{pfy}$

$m_{Mu_{Miy}+ m_{pu_{piy}}+0=(m_{M}+m_{p})V_{f} sin$ Ф

60 ·8 + 0 = (60 + 80) $V_{f}sin$ Ф

480 = 140 $V_{f} sin$ Ф................. (I)

y-axis component form (+y north);

$P_{Mix}$ + $p_{Pix}$ + $j_{x}$ = $P_{Mfx}$ + $P_{pfx}$

$m_{Mu_{Mix}+ m_{pu_{pix}}+0=(m_{M}+m_{p})V_{f} cos$ Ф

0 + 80.9 = (60 + 80) $V_{f}cos$ Ф

720= 140 $V_{f}cos$ Ф

140Vf= $\frac{720}{cos}$ Ф......................................(2)

Substituting (2) into (1) to give the angle;

480 = 720tan Ф

Ф = arctan(0.67) =33.69°.......................(3)

Evaluating (1) with (3) gives the velocity magnitude

480 = 140Vfsin 33.69°

Vf=6.18 m/s

note 1:

This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.

7 0

2 years ago

A small box of mass m1 is sitting on a board of mass m2 and length L (Figure 1) . The board rests on a frictionless horizontal s

chubhunter [2.5K]

Explanation:

Whole system will accelerate under the action of applied force. The box will experience the force against the friction and when this force exceeds then the box will move. so

Ff = μs×m1×g

m1×a = μs×m1×g

a = μs×g

The applied force is given by

F = (m1 + m2)×a so

F = μs×g×(m1+m2)

3 0

1 year ago