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Maslowich
3 years ago
14

A 125-g metal block at a temperature of 93.2 °C was immersed in 100. g of water at 18.3 °C. Given the specific heat of the metal

( s = 0.900 J/g·°C), find the final temperature of the block and the water
Physics
1 answer:
Nataly_w [17]3 years ago
6 0

Answer:

34.17°C

Explanation:

Given:

mass of metal block = 125 g

initial temperature T_i = 93.2°C

We know

Q = m c \Delta T   ..................(1)

Q= Quantity of heat

m = mass of the substance

c = specific heat capacity

c = 4.19 for H₂O in J/g^{\circ}C

\Delta T = change in temperature

Now

The heat lost by metal = The heat gained by the metal

Heat lost by metal = 125\times 0.9\times (93.2-T_f)

Heat gained by the water = 100\times 4.184\times(T_f -18.3)

thus, we have

125\times 0.9\times (93.2-T_f) = 100\times 4.184\times(T_f -18.3)

10485-112.5T_f = 418.4T_f - 7656.72

⇒ T_f = 34.17^oC

Therefore, the final temperature will be = 34.17°C

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300 grams of ethanol is heated with 14640 Joules of energy to reach a final temperature of 30 °C. What was the initial temperatu
levacccp [35]

Answer:

10 °C

Explanation:

Applying

q = cm(t₂-t₁)............... Equation 2

Where q = heat energy, c = specific heat of ethanol, m = mass of ethanol, t₁ = initial temperature, t₂ = Final temperature.

Given: c = 2.44 J/g.°C,  m = 300 g, q = 14640 J, t₂ = 30°C

Substitute into equation 2 and solve for t₁

14640 = 2.44×300(30-t₁)

14640 = 732(30-t₁)

732(30-t₁) = 14640

(30-t₁)  = 14640/732

(30-t₁)  = 20

t₁ = 30-20

t₁ = 10 °C

6 0
2 years ago
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a ball dropped from rest falls freely intil it hits the ground with the speed of 20 m/s . the tine furing which the ball is in f
forsale [732]

Answer:

5m/s

Explanation:

4 0
3 years ago
By what factor is the heat flow increased if a window 0.550 mm on a side is inserted in the door? The glass is 0.450 cmcm, and t
Andrei [34K]

This question is incomplete, the complete question is;

A carpenter builds a solid wood door with dimensions 1.95 m × 0.99 m × 4.5 cm . Its thermal conductivity is k=0.120W/(m.K). The air films on the inner and outer surfaces of the door have the same combined thermal resistance as an additional 1.6 cm thickness of solid wood. The inside air temperature is 19.0°C , and the outside air temperature is -6.50°C .

a) What is the rate of heat flow through the door?

b) By what factor is the heat flow increased if a window 0.550 m on a side is inserted in the door? The glass is 0.450 cm , and the glass has a thermal conductivity of 0.80 W/(m.K). The air films on the two sides of the glass have a total thermal resistance that is the same as an additional 12.0 cm of glass.

Answer:

a) the rate of heat flow through the door is 97 watts

b) The factor of increased heat flow is 1.353

Explanation:

Given that;

room dimension = 1.95m × 0.99m × 4.5cm,

thermal conductivity = 0.120 w/m.k

additional thickness of solid wood Δt = 1.6 cm

a)

first we determine the effective thickness of the door;

t = 4.5cm + 1.6 cm = 61 cm ≈ 0.061 m

Now rate of heat flow is given by the relation

Q = KA( (TH -TC)/L)

= 0.12 × (1.95 m × 0.99 m) × ( (19°C - (-6.50°C)) / 0.061m)

= 0.23166 × 418.0327

= 96.8414 watts

Q = 97 watts

therefore the rate of heat flow through the door is 97 watts

b)

by intensity the glass of thickness 0.450 cm

the effective thickness is

L = 0.45cm + 12 cm = 12.45 cm = 0.1245 m

additionally area of glass A = (0.550 m)²

A = 0.3025 m²

Now

Qglass = KA ((TH-TC)/L)  

= 0.80 w/m.k × 0.3025 m² × (19°C - (-6.50°C)) / 0.1245m)

= 0.242 × 204.819

Qglass = 49.57 watt  

Qwood = KA ((TH-TC)/L)  

area of wooden door = (1.95×0.99) - 0.3025 m² = 1.628m²

so Qwood = 0.12 × 1.628 × (19°C - (-6.50°C)) / 0.061m)

= 0.19536 × 418.0327

Qwood = 81.67 watt

Q = Qglass + Qwood

Q = 49.57 watt  + 81.67 watt

Q = 131.24 watt

The factor of increased heat flow is;

f = 131.24 watt / 97 watts

f = 1.353

8 0
3 years ago
How does the earth orbit the sun?
scoray [572]

Answer:

The Sun's gravity pulls on the planets, just as Earth's gravity pulls down anything that is not held up by some other force and keeps you and me on the ground.

Explanation:

Hope that helps

6 0
3 years ago
Calculate ine gravitational potential energy of the ball using pe=m×g×h.(use g=9.8 n/kg)
Neko [114]

Answer:

58.8J

Explanation:

Given parameters;

Mass of ball  = 4kg

Height above the floor  = 1.5m

g  = 9.8n/kg

Unknown:

Potential energy  = ?

Solution:

The potential energy of a body is the energy due to the position of the body.

It is mathematically expressed as:

  Potential energy = mass x acceleration due to gravity x height

  Potential energy  = 4 x 9.8 x 1.5  = 58.8J

5 0
3 years ago
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