All categories

3 years ago

330 g of water at 55°c are poured into an 855 g aluminum container with an initial temperature of 10°

1 answer:

3 0

The final temperature of the system is 32.5°
we know, H = mcT
where, H = Heat content of the body
m = Mass,
c = Specific heat
T = Change in temperature
According to to the Principle of Calorimetry

The net heat remains constant i.e.
⇒ the heat given by water = heat accepted by the aluminum container.
⇒ 330 x 1 x (45 - T) = 855 x $\frac{900}{4200}$ x (T - 10)

⇒ 14,850 - 330T = 183.21T - 1832

⇒ - 513.21 T = - 16682

or T = 32.5°

You might be interested in

Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudina

Alecsey [184]

(a) 23.4

The fiber-to-matrix load ratio is given by

$\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}$

where

$E_f = 131 GPa$ is the fiber elasticity module

$E_m = 2.4 GPa$ is the matrix elasticity module

$V_f=0.3$ is the fraction of volume of the fiber

$V_m=0.7$ is the fraction of volume of the matrix

Substituting,

$\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4$ (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

We can rewrite (1) as

$F_m = \frac{F_f}{23.4}$

And inserting this into (2):

$F=F_f + \frac{F_f}{23.4}$

Solving the equation, we find the actual load carried by the fiber phase:

$F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N$

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

Using

F = 46500 N

$F_f = 44594 N$

We can immediately find the actual load carried by the matrix phase:

$F_m = F-F_f = 46,500 N - 44,594 N=1,906 N$

(d) 437 MPa

The cross-sectional area of the fiber phase is

$A_f = A V_f$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_f=0.3$ , we have

$A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2$

And the magnitude of the stress on the fiber phase is

$\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa$

(e) 8.0 MPa

The cross-sectional area of the matrix phase is

$A_m = A V_m$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_m=0.7$ , we have

$A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2$

And the magnitude of the stress on the matrix phase is

$\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa$

(f) $3.34\cdot 10^{-3}$

The longitudinal modulus of elasticity is

$E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa$

While the total stress experienced by the composite is

$\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa$

So, the strain experienced by the composite is

$\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}$

3 0

3 years ago

What is the relationship between sunspots and geomagnetic storms?

shutvik [7]

Sunspots are spot like structure on the surface of sun, they are created due to geomagnetic flux which inhibits the convection of heat waves thus reducing the temperature of spot with some factor

8 0

3 years ago

The membrane that surrounds a certain type of living cell has a surface area of 4.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A

Nadusha1986 [10]

Answer:

The charge resides on the outer surface = $1.245 \times 10^{-12}$ C

Explanation:

Surface area of cell $(A) = 4.3\times 10^{-9} m^{2}$

Separation between two plate $(d) = 1.1 \times 10^{-8} m$

Dielectric constant $(k) = 4.2$

Potential difference $(\Delta V) = 85.7 \times 10^{-3} V$

The capacitance of parallel plate capacitor in free space is given by,

$C = \frac{\epsilon_{o} A }{d}$

Where $\epsilon_{o} =$ permittivity of free space = $8.85 \times 10^{-12}$

The Capacitance of capacitor is increase by $k$ times when it placed in dielectric medium.

$C_{dielectric} = \frac{k \epsilon_{o} A }{d}$

And we know that, $C = \frac{Q}{ \Delta V}$

So charge on the outer surface is given by,

$Q = \frac{k \epsilon A \Delta V }{d}$

$Q = \frac{4.2 \times 4.3 \times 10^{-9} \times 8.85 \times 10^{-12} \times 85.7\times 10^{-3} }{1.1 \times 10^{-8} }$

$Q = 1.245 \times 10^{-12}$

3 0

3 years ago

How death rate helps in changing population

lapo4ka [179]

Both the birth and death rate are expressed per 1000 of the population.

6 0

3 years ago

A car is running at a velocity of 50 miles/hour and the driver accelerates the car by 10miles/hour.How far the car travels from

dezoksy [38]

Initial velocity u = 50 miles/hour
acceleration a = 10 miles/hour
Time t = 2 hours
Distance travelled S = ut + (at^2)/2
Substituting the values in the second equation of motion,
S = 50*2 + (10 * 2 *2)/2
S = 100 + 20
S = 120 miles
Therefore the distance travelled by the car in the next two hours is 120 miles

4 0

3 years ago

Read 2 more answers