True or false

Part A What will be the equilibrium temperature when a 227 g block of copper at 283 °C is placed in a 155 g aluminum calorimeter

stellarik [79]

Answer:

T = 20.84°C

Explanation:

From the law of conservation of energy:

Heat Lost by Copper Block = Heat Gained by Aluminum Calorimeter + Heat Gained by Water

$m_cC_c\Delta T_c = m_wC_w\Delta T_w + m_aC_a\Delta T_a$

where,

$m_c$ = mass of copper = 227 g

$m_w$ = mass of water = 844 g

$m_a$ = mass of aluminum = 155 g

$C_c$ = specific heat capacity of calorimeter = 385 J/kg.°C

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

$C_a$ = specific heat capacity of aluminum = 890 J/kg.°C

$\Delta T_c$ = change in temperature of copper = 283°C - T

$\Delta T_w$ = change in temperature of water = T - 14.6°C

$\Delta T_a$ = change in temperature of aluminum = T - 14.6°C

T = equilibrium temperature = ?

Therefore,

$(227\ g)(385\ J/kg.^oC)(283^oC-T)=(844\ g)(4200\ J/kg.^oC)(T-14.6^oC)+(155\ g)(890\ J/kg.^oC)(T-14.6^oC)\\\\24732785\ J - (87395\ J/^oC) T = (3544800\ J/^oC) T - 51754080\ J+ (137950\ J/^oC) T-2014070\ J\\\\24732785\ J +51754080\ J+2014070\ J = (3544800\ J/^oC) T+(137950\ J/^oC+(87395\ J/^oC) T\\\\78560935\ J = (3770145\ J/^oC) T\\\\T = \frac{78560935\ J}{3770145\ J/^oC}$

8 0

2 years ago

What causes some materials to have a magnetic fields?

Viktor [21]

Well it determines on how much magnetism a material has.

6 0

3 years ago

Read 2 more answers

The work function (energy needed to remove an electron) of gold is 5.1 eV. Two pieces of gold (at the same potential) are separa

scoundrel [369]

Answer:

The value of L is 0.3 nm.

Explanation:

Given that,

Energy $\phi= 5.1 eV$

Transmission probability = 10⁻³

We need to calculate the value of L

We know that,

Formula of tunneling probability

$T=Ge^{-2kL}$ ....(I)

Where,

$k=\dfrac{\sqrt{2m(E-U)}}{\hbar}$ ....(II)

Where, m = mass of electron

$E = \phi+U$

$\phi = E-U$

Put the value in equation (II)

$k=\dfrac{\sqrt{2\times9.1\times10^{-31}\times5.1\times1.6\times10^{-19}}}{1.055\times10^{-34}}$

$k=1.155\times10^{10} m^{-1}$

From equation (I)

$ln T=-2kLG$

$L=\dfrac{ln T}{-2kG}$

$L=\dfrac{ln 10^{-3}}{-2\times1.155\times10^{10}\times1}$

$L=2.99\times10^{-10}\ m$

$L=0.299\times10^{-9}\ m$

$L=0.3\ nm$

Hence, The value of L is 0.3 nm.

7 0

3 years ago

A load of 20N is lifted through a height of 8m by a machine.If the effort of 80N moves a distance of 6m.Calculate the efficiency

I am Lyosha [343]

A load of 20N is lifted through a height of 8m by a machine.If the effort of 80N moves a distance of 6m.Calculate the efficiency percentage of the machine

Explanation:

A load of 20N is lifted through a height of 8m by a machine.If the effort of 80N moves a distance of 6m.Calculate the efficiency percentage of the machine

8 0

2 years ago

Read 2 more answers

A 50 km/hr wind is blowing due northwest what are the components of the wind in the northern

jeka57 [31]

Answer:

The components of the wind in the northern direction is approximately 35.36 km/hr

Similarly, the components of the wind in the western direction is approximately 35.36 km/hr

Explanation:

The given parameters are;

The magnitude of the speed of the wind, v = 50 km/hr

The direction of the wind = The northwest direction

Therefore;

Given that the angle between the northern and western direction = 90°

The angle in the northwestern direction, θ = 90°/2 = 45°

Therefore;

The components of the wind in the northern direction = $v_y$ = v × sin(θ)

Substituting the values, we have;

The components of the wind in the northern direction = $v_y$ = 50 × sin(45°) = 50/√2 ≈ 35.36 km/hr.

Similarly, the components of the wind in the western direction = vₓ = 50 × cos(45°) = 50/√2 ≈ 35.36 km/hr

8 0

2 years ago