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3 years ago

What would the final temperature be if 8.94 X 10 3 joules of heat were added to 454 grams of copper at 23.0 o C?

Physics

1 answer:

ozzi3 years ago

8 0

Answer: $74.1^0C$

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= $8.94\times 10^3$ Joules

m= mass of copper = 454 g

c = specific heat capacity = $0.385J/g^0C$

Initial temperature of the copper = $T_i$ = 23.0°C

Final temperature of the water = $T_f$ = ?

Change in temperature , $\Delta T=T_f-T_i$

Putting in the values, we get:

$8.94\times 10^3=454\times 0.385\times (T_f-23.0)^0C$

$T_f=74.1^0C$

The final temperature of copper will be $74.1^0C$

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3 years ago

an empty density bottle weighed 23.5 and 48.4g when filled with water .The empty bottle was partially filled with sand until and

gavmur [86]

Answer:

Density of Sand is $2.653g/cm^{3}$ .

Explanation:

Given Empty Density bottle weighs 23.5gm(W=23.5gm)

Weight of bottle when completely filled water=48.4gm

So amount of water required to fill the bottle=Weight of bottle filled with water-W

Amount of water required to fill the bottle( $w_{max}$ )=48.4gm-23.5gm

$w_{max}=24.9$ g

Since we know density of water $d_{w} =1g/cm^{3}$ and $w_{max}$

We can calculate volume of empty space in the bottle(V).

$w_{max}=d_{w}$ $\times$ V

V= $\frac{w_{max} }{d_{w} }$

V= $\frac{24.9}{1}$

V=24.9 $g/cm^{3}$

Now bottle is partially filled with sand,and weight of bottle is ( $w_{s}$ )36.5gm

So,

Amount of sand added ( $m_{s}$ )=36.5-Weight of the bottle

$m_{s}$ =13g

After filling the bottle with water again,the weight of the bottle becomes ( $W_{2}$ =56.5g)

Therefore,

amount of water added to the bottle of sand in grams = $W_{2}$ -36.5gm

amount of water added =56.5g-36.5g

amount of water added =20g

As the density of water = 1g/ $cm^{3}$

Amount of water (in grams )=Volume of water occupied

20=volume of water added

Therfore volume of water added to the sand filled bottle( $V_{w}$ )=20 $cm^{3}$

As we know the total volume of the water bottle(V),

Volume of the sand occupied in the water bottle=V- $V_{w}$

$V_{s}$ =24.9g-20g

$V_{s}$ =4.9g

We know,

Density=Mass/Volume

Therefore,

density of sand = $\frac{m_{s} }{V_{s} }$

density of sand = $\frac{13}{4.9}$

density of sand = $2.653g/cm^{3}$

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3 years ago

How much energy will be transferred to your eardrum while listening to this sound for 1.0 min?

Phoenix [80]

So E = 2x10^-3W/m^2*(π*(3.0x10^-3m)^2)*1min*60s... = 3.4x10^-6J

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3 years ago

At the end of the Shock Wave roller coaster ride, the 6000-kg train of cars (includes passengers) is allowed from a speed of 20

Angelina_Jolie [31]

Answer:

56250 N

Explanation:

mass, m = 6000 kg

initial speed, u = 20 m/s

final speed, v = 5 m/s

distance, s = 20 m

Use third equation of motion

$v^{2}=u^{2}+2as$

5 x 5 = 20 x 20 + 2 a x 20

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a = - 9.375 m/s^2

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F = 6000 x 9.375

F = 56250 N

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3 years ago

Two large metal plates of area 0.88 m2 face each other, 4.8 cm apart, with equal charge magnitudes but opposite signs. The field

Nuetrik [128]

Answer:

q=6.22*10^-10C

Explanation:

Two large metal plates of area 0.88 m2 face each other, 4.8 cm apart, with equal charge magnitudes but opposite signs. The field magnitude E between them (neglect fringing) is 80 N/C. Find |q|

E=α/∈, electric field within the plate

α=q/A

A=area of the plate

∈=is the permittivity

substituting , we have

The field magnitude E between them (neglect fringing)

E=q/A∈

q=EA∈

q=0.88*80*8.84*10^-12

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3 years ago