Explain what an electric current could do if applied to a pickle!!.....<br><br> I’m waiting........

An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10

Alexeev081 [22]

Answer:

a) $c=1822.3214\ J.kg^{-1}.K^{-1}$

b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c) The material is peat, possibly.

d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.

Explanation:

Given:

mass of aluminium, $m_a=0.1\ kg$

mass of water, $m_w=0.25\ kg$

initial temperature of the system, $T_i=10^{\circ}C$

mass of copper block, $m_c=0.1\ kg$

temperature of copper block, $T_c=50^{\circ}C$

mass of the other block, $m=0.07\ kg$

temperature of the other block, $T=100^{\circ}C$

final equilibrium temperature, $T_f=20^{\circ}C$

We have,

specific heat of aluminium, $c_a=910\ J.kg^{-1}.K^{-1}$

specific heat of copper, $c_c=390\ J.kg^{-1}.K^{-1}$

specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

Using the heat energy conservation equation.

The heat absorbed by the system of the calorie-meter to reach the final temperature.

$Q_{in}=m_a.c_a.(T_f-T_i)+m_w.c_w.(T_f-T_i)$

$Q_{in}=0.1\times 910\times (20-10)+0.25\times 4186\times (20-10)$

$Q_{in}=11375\ J$

The heat released by the blocks when dipped into water:

$Q_{out}=m_c.c_c.(T_c-T_f)+m.c.(T-T_f)$

where

$c=$ specific heat of the unknown material

For the conservation of energy : $Q_{in}=Q_{out}$

so,

$11375=0.1\times 390\times (50-20)+0.07\times c\times (100-20)$

$c=1822.3214\ J.kg^{-1}.K^{-1}$

b)

This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c)

The material is peat, possibly.

d)

The material cannot be ice because ice doesn't exists at a temperature of 100°C.

7 0

3 years ago

Imagine that while you and a passenger are in a deep-diving submersible in the North Pacific near Alaska’s Aleutian Islands, you

mamaluj [8]

Answer:

I would say its a deep ocean trench

Explanation:

This is because deep ocean trenches are found at the deepest part of the ocean and also at Pacific ocean margins or Rim where subduction usually occurs and Aleutian islands are part of the Pacific Rim

6 0

3 years ago

Astronomers know that the distance between the Earth and the Sun averages 1.50 x108 km. How can astronomers use the observed ste

rodikova [14]

Answer:

The distance of stars and the earth can be averagely measured by using the knowledge of geometry to estimate the stellar parallax angle(p).

From the equation below, the stars distances can be calculated.

D = 1/p

Distance = 1/(parallax angle)

Stellar parallax can be used to determine the distance of stars from an observer, on the surface of the earth due to the motion of the observer. It is the relative or apparent angular displacement of the star, due to the displacement of the observer.

Explanation:

Parallax is the observed apparent change in the position of an object resulting from a change in the position of the observer. Specifically, in the case of astronomy it refers to the apparent displacement of a nearby star as seen from an observer on Earth.

The parallax of an object can be used to approximate the distance to an object using the formula:

D = 1/p

Where p is the parallax angle observed using geometry and D is the actual distance measured in parsecs. A parsec is defined as the distance at which an object has a parallax of 1 arcsecond. This distance is approximately 3.26 light years

3 0

3 years ago

The intensity I of light varies inversely as the square of the distance D from the source. If the intensity of illumination on a

emmasim [6.3K]

The intensity on a screen 20 ft from the light will be 0.125-foot candles.

<h3>What is the distance?</h3>

Distance is a numerical representation of the length between two objects or locations.

The intensity I of light varies inversely as the square of the distance D from the source;

I∝(1/D²)

The ratio of the intensity of the two cases;

$\rm \frac{I_1}{I_2} =(\frac{D_2}{D_1} )^2\\\\ \rm \frac{2}{I_2} =(\frac{20}{5} )^2\\\\ \frac{2}{I_2} =4^2 \\\\ I_2= \frac{2}{16} \\\\ I_2= 0.125 \ foot-candles$