Which of the following is a heterogeneous mixture? A. salt B. dye in water C. sugar water D. a garden salad

You have to heat up a 1.000 gram ingot of aluminum from initial temperature Ti = 847 K to its melting point, 933 K. Calculate th

alexdok [17]

Answer: 18.9 calories

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat absorbed

m= mass of substance = 1.0 g = 0.001 kg

c = specific heat capacity of aluminium = $921.096J/kgK$

Initial temperature = $T_i$ = 847 K

Final temperature = $T_f$ = 933 K

Change in temperature , $\Delta T=T_f-T_i=(933-847)K=86K$

Putting in the values, we get:

$Q=0.001kg\times 921.096J/kgK\times 86K$

$Q=79.2J=18.9cal$ (1J=0.24cal)

The number of calories required for this heat up process is 18.9 calories.

3 0

3 years ago

A child has two red wagons, with the rear one tied to the front by a (non-stretching) rope. If the child pushes on the rear wago

Sergeeva-Olga [200]

Answer:

The rear wagon gains the kinetic energy, but the front wagon will remain at rest.

The two-wagon system will gain a kinetic energy $\dfrac{1}{16}$ of the kinetic energy gained by the rear wagon.

Explanation:

Let's consider that the masses of the wagons to be 'M'. When the child pushes the rear wagon let's assume that the velocity of the rear wagon be 'v'.

Therefor the kinetic energy gained by the rear wagon be $K_{r} = \frac{1}{2}Mv^{2}$ .

Now let's assume that the velocity of the centre of mass (C), as shown in the figure, be 'V'. So from momentum conservation law we can write,

$&& Mv = (M + M)V\\\\&or,& V = \frac{v}{2}$

Now the centre of mass ( $M_{C}$ ) is given by

$M_{C} = \frac{M \times M}{M + M} = \frac{M}{2}$

So the kinetic energy ( $K_{C}$ ) of the system will be

$K_{C} = \frac{1}{2}M_{C}V^{2} = \frac{1}{2}\frac{M}{2}(\frac{v}{2})^{2} = \frac{1}{16}Mv^{2}$

5 0

3 years ago

Gasoline flows in a long, underground pipeline at a constant temperature of 15o C (rho = 680 kg/m3 ; ν = 4.6 × 10-7 m2 /s). Two

poizon [28]

Answer:

1.0416 m∧3/sec

Explanation:

check the pictures below for the solution

8 0

3 years ago

Steam undergoes an adiabatic expansion in a piston–cylinder assembly from 100 bar, 360°C to 1 bar, 160°C. What is work in kJ per

vfiekz [6]

Answer:

work is 130.5 kJ/kg

entropy change is 1.655 kJ/kg-k

maximum theoretical work is 689.4 kJ/kg

Explanation:

piston cylinder assembly

100 bar, 360°C to 1 bar, 160°C

to find out

work and amount of entropy and magnitude

solution

first we calculate work i.e heat transfer - work = specific internal energy @1 bar, 160°C - specific internal energy @ 100 bar, 360°C .................1

so first we get some value from steam table with the help of 100 bar @360°C and 1 bar @ 160°C

specific volume = 0.0233 m³/kg

specific enthalpy = 2961 kJ/kg

specific internal energy = 2728 kJ/kg

specific entropy = 6.004 kJ/kg-k

and respectively

specific volume = 1.9838 m³/kg

specific enthalpy = 2795.8 kJ/kg

specific internal energy = 2597.5 kJ/kg

specific entropy = 7.659 kJ/kg-k

now from equation 1 we know heat transfer q = 0

so - w = specific internal energy @1 bar, 160°C - specific internal energy @ 100 bar, 360°C

work = 2728 - 2597.5

work is 130.5 kJ/kg

and entropy change formula is i.e.

entropy change = specific entropy ( 100 bar @360°C) - specific entropy ( 1 bar @160°C )

put these value we get

entropy change = 7.659 - 6.004

entropy change is 1.655 kJ/kg-k

and we know maximum theoretical work = isentropic work

from steam table we know specific internal energy is 2038.3 kJ/kg

maximum theoretical work = specific internal energy - 2038.3

maximum theoretical work = 2728 - 2038.3

maximum theoretical work is 689.4 kJ/kg

3 0

3 years ago

Which is not a factor that affects the pressure of a gas in a closed container?

maksim [4K]

it is size of of particles because it does not matter about the size in a closed container

7 0

3 years ago