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3 years ago

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How much electricity is used to boil 600 g of water if the kettle has a power of 1500 W? The water boiled for 3 minutes and 9 se

conds. Water density is 1000 kg/m3, specific heat of water is 4200 J/(kg· oC).

1 answer:

Ksju [112]3 years ago

3 0

Answer:

The electrical energy consumed in boiling the water is 0.0788 kWh

Explanation:

Given;

mass of the water, m = 600 g = 0.6 kg

power rating of the kettle, P = 1500 W = 1.5 kW

specific heat capacity of water, c = 4,200 J/kg⁰C

density of water, = 1000 kg/m³

time taken to boil the water, t = 3 mins + 9 s = (3 x 60s) + 9 s = 180 s + 9 s = 189 s = $189 \ s \times \frac{1 hr}{3.600 \ s} = 0.0525 \ hr$

The electrical energy consumed in boiling the water is calculated as;

E = P x t

E = 1.5 kW x 0.0525 hr

E = 0.0788 kWh

Therefore, the electrical energy consumed in boiling the water is 0.0788 kWh

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2) A Ship has an area of cross - section at the water line of 2000m²

STatiana [176]

Answer:

a. 2 m

b. 0.15 m

Explanation:

(a) By what deck does the ship sink in fresh.

water, when it loads a cargo of 4000 tonnes

We know that the upward force , U on the ship equal the weight of fresh water displaced, W = ρVg where ρ = density of fresh water = 1000 kg/m³, V = volume of water displaced and g = acceleration due to gravity.

So, U = W = ρVg

This upward force must equal the weight of the ship, W' = mg where m = mass of ship = 4000 tonnes = 4000 × 1000 kg = 4 × 10⁶ kg

So. U = W'

ρVg = mg

V = m/ρ = 4 × 10⁶ kg/10³ kg/m³ = 4 × 10³ m³

Now the volume of water displaced equals the volume occupied by the ship floating ship of cross-sectional area, A = 2000 m³ and depth in water h.

So, V = Ah

Thus h = V/A = 4 × 10³ m³/2000 m³ = 2 m

So, the ship sinks to a depth of 2 m in fresh water.

(b) if the ship + Cargo has a displacement tonnage

of 12300 tonnes; by what amount will the ship

rise in the water when it sails from fresh water

into Seawater (density of Sea water - 1025kgm⁻³

We know that the upward force , U' on the ship in fresh water equals the weight of fresh water displaced, W" = ρV'g where ρ = density of fresh water = 1000 kg/m³, V' = volume of water displaced and g = acceleration due to gravity.

So, U = W" = ρV'g

This upward force must equal the weight of the ship + cargo, W₁ = m₁g where m₁ = mass of ship + cargo = 12300 tonnes = 12300 × 1000 kg = 1.23 × 10⁷ kg

So. U' = W₁

ρV'g = m₁g

V' = m₁/ρ = 1.23 × 10⁷ kg/10³ kg/m³ = 1.23 × 10⁴ m³

Now the volume of water displaced equals the volume occupied by the ship floating ship of cross-sectional area, A = 2000 m³ and depth in fresh water h'.

So, V' = Ah'

Thus h = V'/A = 1.23 × 10⁴ m³/2000 m³ = 6.15 m

So, the ship sinks to a depth of 0.6 m in fresh water.

Also,

We know that the upward force , U" on the ship in sea water equals the weight of sea water displaced, W₂ = ρ'V"g where ρ' = density of sea water = 1025 kg/m³, V"= volume of water displaced and g = acceleration due to gravity.

So, U" = W₂ = ρ'V"g

This upward force must equal the weight of the ship + cargo, W₁ = m₁g where m₁ = mass of ship + cargo = 12300 tonnes = 12300 × 1000 kg = 1.23 × 10⁷ kg

So. U" = W₁

ρ'V"g = m₁g

V" = m₁/ρ' = 1.23 × 10⁷ kg/1.025 × 10³ kg/m³ = 1.2 × 10⁴ m³

Now the volume of water displaced equals the volume occupied by the ship floating ship of cross-sectional area, A = 2000 m³ and depth in sea water h".

So, V" = Ah"

Thus h" = V'/A = 1.2 × 10⁴ m³/2000 m³ = 6 m

So, the ship sinks to a depth of 6 m in fresh water.

So, the rise in height from fresh water to sea water is Δh = h' - h" = 6.15 m - 6 m = 0.15 m

6 0

3 years ago

Consider two closed systems A and B. System A contains 3000 kJ of thermal energy at 20°C, whereas system B contains 200 kJ of th

Soloha48 [4]

No because they’re closed.

4 0

3 years ago

A 33.0-kg child starting from rest slides down a water slide with a vertical height of 15.0 m. (Neglect friction.)

Temka [501]

This question involves the concept of the conservation of energy.

(a) The child's speed halfway will be "12.13 m/s".

(b) The child's speed three-fourth way will be "14.86 m/s".

(a)

Using the <em>law of conservation of energy</em>:

Potential Energy Lost = Kinetic Energy Gained

$mgh = \frac{1}{2}mv^2\\\\2gh = v^2\\\\v = \sqrt{2gh}$

where,

v = speed = ?

g = acceleration due to gravity = 9.81 m/s²

h = height lost = halfway = height/2 = 15 m/2 = 7.5 m

Therefore,

$v = \sqrt{(2)(9.81\ m/s^2)(7.5\ m)}$

<u>v = 12.13 m/s</u>

<u />

(b)

Using the <em>law of conservation of energy</em>:

Potential Energy Lost = Kinetic Energy Gained

$mgh = \frac{1}{2}mv^2\\\\2gh = v^2\\\\v = \sqrt{2gh}$

where,

v = speed = ?

g = acceleration due to gravity = 9.81 m/s²

h = height lost = three-fourth way down = $\frac{3}{4}height = \frac{3}{4}(15\ m) = 11.25\ m$

Therefore,

$v = \sqrt{(2)(9.81\ m/s^2)(11.25\ m)}$

<u>v = 14.86 m/s</u>

<u />

Learn more about the <em>law of conservation</em> of energy here:

brainly.com/question/20971995?referrer=searchResults

The attached picture shows the <em>law of conservation of energy</em>.

7 0

3 years ago

One and one-half moles of an ideal monatomic gas expand adiabatically, performing 7900 J of work in the process. What is the cha

mart [117]

Answer:

The temperature change is -633.15K

Explanation:

If we considered the expansion as a reversible one (to be adiabatic is one of the requirements), the work done by expansion can be written as:

$W=-n*P(v_2-v_1)=-P*(V_2-V_1)$ Where 2 and 1 subscripts mean the final and the initial state respectively. The equation negative sign says that for an expansion of the gas, the system is making work, so the energy is going out of the system.

Using the ideal gas equation, it is possible to change volume and pressure by temperatures:

$PV_2=(nRT_2)\\PV_1=nRT_1$

So,

$W=-nR(T_2-T_1)$

$(T_2-T_1)=\frac{-W}{nR}=\frac{-7900J}{(1.5mol*8.314\frac{J}{molK})}=-633.5K$

This result makes sense considering that the volume increases, so it is expected that the temperature decreases.

4 0

3 years ago

a 45kg boy is holding a 12kg pumpkin while standing on ice skates on a smooth frozen pond. the boy tosses the pumpkin with a hor

irga5000 [103]

Let's split the problem into two parts, and let's solve it by using conservation of momentum.

1) Let's analyze the moment when the boy tosses the pumpkin. Before this moment, the boy is standing with the pumpkin, so with speed v=0 and therefore the total momentum of the system (boy+pumpkin) is zero.
After he tosses the pumpkin, the boy will start to move with speed $v_b$ and the pumpkin starts to move with speed $v_p=2.8 m/s$ . The momentum of the boy is $m_b v_b$ (with $m_b =45 kg$ being the mass of the boy), while the momentum of the pumpkin is $m_p v_p$ (where $m_p=12 kg$ is the mass of the pumpkin). Since the total momentum must be equal to zero (because the total momentum cannot change), then we can write
$m_b v_b + m_p v_p =0$
From which we find
$v_b = - \frac{m_p v_p}{m_b}=- \frac{(12 kg)(2.8 m/s)}{45 kg} =-0.75 m/s$
and this is the speed of the boy, and the negative sign means he's moving in the opposite direction of the pumpkin.

2) Now let's focus on the entire system (boy+pumpkin+girl). Initially, the total momentum of this system is zero, because both the boy (holding the pumpkin) and the girl are still. So, the total momentum after the girl catches the pumpkin must be still zero.
After this moment, the boy has a momentum of $m_b v_b$ , while the girl has momentum $(m_g+m_p)v_g$ where $m_g=37 kg$ is the girl mass $v_g$ is the girl speed. Here we use $m_g+m_p$ because the girl is holding the pumpkin now. Therefore, the conservation of momentum becomes
$m_bv_b + (m_g+m_p)v_g =0$
and so
$v_g = - \frac{m_b v_b}{m_g+m_p} =- \frac{(45 kg)(-0.75 m/s)}{37 kg+12 kg} =0.69 m/s$
and this is the girl's speed, with positive sign so with same direction of the pumpkin initial direction.

6 0

4 years ago