Ask question

Ask question

All categories

mihalych1998 [28]

3 years ago

8

You place a piece of aluminum at 250.0∘C ∘ C in 9.00 kg k g of liquid water at 20.0∘C ∘ C . None of the water boils, and the fin

al temperature of the water and aluminum is 22.0∘C ∘ C . What is the mass of the piece of aluminum? Assume no heat is exchanged with the container that holds the water.

1 answer:

Mice21 [21]3 years ago

5 0

Answer:

m₁ = 0.37 kg

Explanation:

According to Law of conservation of energy:

Heat Lost by Aluminum = Heat Gained by Water

m₁C₁ΔT₁ = m₂C₂ΔT₂

where,

m₁ = mass of piece of aluminum = ?

C₁ = specific heat capacity of aluminum = 900 J/kg.°C

ΔT₁ = Change in temperature of aluminum = 250°C - 22°C = 228°C

m₂ = mass of water = 9 kg

C₂ = specific heat capacity of water = 4200 J/kg.°C

ΔT₁ = Change in temperature of aluminum = 22°C - 20°C = 2°C

Therefore,

m₁(900 J/kg.°C)(228 °C) = (9 kg)(4200 J/kg.°C)(2°C)

m₁ = (75600 J)/(205200 J/kg)

<u>m₁ = 0.37 kg</u>

You might be interested in

A particular automotive wheel has an angular moment of inertia of 12 kg*m^2, and is decelerated from 135 rpm to 0 rpm in 8 secon

lozanna [386]

Answer:

(A) Torque required is 21.205 N-m

(b) Wok done will be equal to 1199.1286 j

Explanation:

We have given moment of inertia $I=12kgm^2$

Wheel deaccelerate from 135 rpm to 0 rpm

135 rpm = $135\times \frac{2\pi }{60}=14.1371rad/sec$

Time t = 8 sec

So angular speed $\omega _i=135rpm$ and $\omega _f=0rpm$

Angular acceleration is given by $\alpha =\frac{\omega _f-\omega _i}{t}=\frac{0-14.1371}{8}=--1.7671rad/sec^2$

Torque is given by torque $\tau =I\alpha$

$=12\times 1.7671=21.205N-m$

Work done to accelerate the vehicle is

$\Delta w=K_I-K_F$

$\Delta W=\frac{1}{2}\times 12\times 14.137^2-\frac{1}{2}\times 12\times0^2=1199.1286J$

8 0

4 years ago

wo skaters collide and embrace in an inelastic collision. Alex's mass is 90 kg and his initial velocity is 1.5 m/s i . Barbara's

Natalija [7]

Answer:

The two skaters will move with a common speed of 1.19 m/s.

Explanation:

Given that,

Mass of Alex, $m_1=90\ kg$

Initial velocity of Alex, $u_1=1.5i\ m/s$

Mass of Barbara, $m_2=57\ kg$

Initial velocity of Barbara, $u_2=2j\ m/s$

After the collision, the two skaters move together at a common velocity. Let V is the common velocity. Using the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V\\\\90\times 1.5i+57\times 2j=(90+57)V\\\\V=\dfrac{135i+114j}{147}\\\\V=\dfrac{135i}{147}+\dfrac{114j}{147}\\\\V=(0.91i+0.77j)\ m/s$

Magnitude of final velocity:

$|V|=\sqrt{0.91^2+0.77^2} \\\\|V|=1.19\ m/s$

So, the two skaters will move with a common speed of 1.19 m/s.

3 0

4 years ago

Read 2 more answers

An object is taken from an oven at 350o F and left to cool in a room at 70o F. If the temperature fell to 250o F in one hour, wh

Oksana_A [137]

Answer:

117.83° F

Explanation:

Using Newton's Law of Cooling which can be expressed as:

$\dfrac{dT}{dt}= k(T-T_1)$

The differential equation can be computed as:

$\dfrac{dT}{dt}= k(T-70)$

$\dfrac{dT}{(T-70)}= kdt$

$\int \dfrac{dT}{(T-70)}= \int kdt$

$In|T-70| = kt +C$

$T- 70 = e^{kt+C} \\ \\ T = 70+e^{kt+C} \\ \\ T = 70 + C_1e^{kt} --- (1)$

where;

$C_1 = e^C$

At the initial condition, T(0)= 350

$350 = 70 C_1^{k*0}$

$350 -70 = C_1$

$280 = C_1$

replacing $C_1$ = 280 into (1)

Hence, the differential equation becomes:

$T(t) = 70 + 280 e^{kt}$

when;

time (t) = 1 hour

T(1) = 250

Since;

$250 = 70 + 280 e^{k*1}$

$180 = 280e^k \\ \\ \dfrac{180}{280}= e^k$

$k = In (\dfrac{180}{280})$

k = -0.4418

Therefore;

T(t) = 70 + 280e^{(-0.4418)}t

After 4 hours, the temperature is:

T(t) = 70 + 280e^{(-0.4418)}4

T(4) = 117.83° F

7 0

3 years ago

ANSWER = BRAINLIEST<br><br> Attachment included ...

musickatia [10]

The answer is a 16000kg

5 0

3 years ago

A parallel-plate capacitor is connected to a battery. What happens to the stored energy if the plate separation is doubled while

zlopas [31]

Answer:

new energy of the capacitor is half that of initial energy

Explanation:

As we know that the energy stored in the capacitor is given as

$U = \frac{1}{2}CV^2$

here we know that capacitance of parallel plate capacitor is given as

$C = \frac{\epsilon_0 A}{d}$

V = voltage of battery

so now we have capacitor remains connected to the same battery and the separation between the plates is doubled

so we have

$C' = \frac{\epsilon_0 A}{2d}$

so now the energy stored between the plates is

$U = \frac{1}{2}(\frac{C}{2})V^2$

so new energy of the capacitor is half

8 0

4 years ago