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3 years ago

How much heat is absorbed by a 29g iron skillet when its temperature rises from 9oC to 105oC?

1 answer:

5 0

We have:

Mass (m) of the iron skillet = 29 grams
Difference in temperature (Δt) = 105 - 9 = 96°C
Heat capacity of iron = 0.449 J/g °C

The energy (Q) is given by:

Q = m × Δt × specific heat
Q = 29 grams × 96 °C × 0.449 J/g°C
Q = 1250.016 Joules

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A recent home energy bill indicates that a household used 475475 kWh (kilowatt‑hour) of electrical energy and 135135 therms for

faltersainse [42]

Answer:

Explanation:

Given that,.

A house hold power consumption is

475 KWh

Gas used is

135 thermal gas for month

Given that, 1 thermal = 29.3 KWh

Then,

135 thermal = 135 × 29.3 = 3955.5 KWh

So, total power used is

P = 475 + 3955.5

P =4430.5 KWh

Since 1 hr = 3600 seconds

So, the energy consumed for 1hr is

1KW = 1000W

P = energy / time

Energy = Power × time

E = 4430.5 KWhr × 1000W / KW × 3600s / hr

E = 1.595 × 10^10 J

So, using Albert Einstein relativity equation

E = mc²

m = E / c²

c is speed of light = 3 × 10^8 m/s

m = 1.595 × 10^10 / (3 × 10^8)²

m = 1.77 × 10^-7 kg

Then,

1 kg = 10^6 mg

m = 1.77 × 10^-7 kg × 10^6 mg / kg

m = 0.177mg

m ≈ 0.18 mg

5 0

3 years ago

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

Elanso [62]

Answer:

The value is $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

The initial speed is $u = 2.05 *10^{4} \ m/s$

Generally the total energy possessed by the space probe when on earth is mathematically represented as

$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

=> $KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2$

=> $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=> $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$

Generally from the law energy conservation we have that

$T__{E}} = KE_p$

$2.101 *10^{8} m + 6.272 *10^{7} m = \frac{1}{2} * m * v^2$

=> $5.4564 *10^{8} = v^2$

=> $v = \sqrt{5.4564 *10^{8}}$

=> $v = 2.3359 *10^{4} \ m/s$

4 0

2 years ago

PLEASE HELP!!!!! (20 points!)

valentina_108 [34]

Step 2
step 5
step 4
step 6
step 1
step 3

thank me later

3 0

3 years ago

Read 2 more answers

A heat pump with a COP of 3.15 is used to heat an air-tight house. When running, the heat pump consumes 5 kW of power. If the te

Jet001 [13]

Answer: 1026s, 17.1m

Explanation:

Given

COP of heat pump = 3.15

Mass of air, m = 1500kg

Initial temperature, T1 = 7°C

Final temperature, T2 = 22°C

Power of the heat pump, W = 5kW

The amount of heat needed to increase temperature in the house,

Q = mcΔT

Q = 1500 * 0.718 * (22 - 7)

Q = 1077 * 15

Q = 16155

Rate at which heat is supplied to the house is

Q' = COP * W

Q' = 3.15 * 5

Q' = 15.75

Time required to raise the temperature is

Δt = Q/Q'

Δt = 16155 / 15.75

Δt = 1025.7 s

Δt ~ 1026 s

Δt ~ 17.1 min

5 0

3 years ago

Read 2 more answers

Which process can transfer energy to a solid metal block

gladu [14]

Eneregy is the ability to to do work it takes work to heat someone heat is energy put the solid metal block in a campfire and heat it up tada wait, we need the options, nevermind

4 0

3 years ago

Read 2 more answers