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alisha [4.7K]
3 years ago
13

How much heat is lost by 2.0 grams of water if the temperature drops from 31 °C to 29 °C? The specific heat of water is 4.184 J/

(g • °C) *
1 point

a) 4.0 J

b) - 7.5 J

c)- 8.4 J

d) - 16.7 J​
Physics
1 answer:
Elanso [62]3 years ago
4 0

Given :

Mass of water, m = 2 grams.

The temperature of water drops from 31 °C to 29 °C .

The specific heat of water is 4.184 J/(g • °C).

To Find :

Amount of heat lost in this process.

Solution :

We know, heat lost is given by :

Heat\ lost,H = ms( T_f - T_i)\\\\H = 2\times 4.184 \times ( 31 - 29 )\ J\\\\H = 16.736\ J

Therefore, amount of heat lost in this process is 16.736 J.

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2 years ago
I will pick the branliest if answered these questions.
Bond [772]

1.

n₁ = index of refraction of diamond = 2.4

n₂ = index of refraction of water = 1.33

θ₁ = angle of incidence = 24 deg

θ₂ = angle of refraction = ?

using snell's law

n₁ Sinθ₁ = n₂ Sinθ₂

inserting the values

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2.

c = specific heat = 4.18

m = mass of water = 3.5 kg

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Q = heat taken

heat taken is given as

Q = m c ΔT

inserting the values

Q = (3.5) (4.18) (30)

Q = 439 J


3.

n = number of moles = 1

m = molar mass = 0.0399 kg

T = temperature = 27 K

v = average velocity

average velocity is given as

v = sqrt(3RT/m)

v = sqrt(3 x 8.314 x 275/0.0399)

v = 414.6m/s


4.

n = number of moles = 2

T = temperature = 35 C  = 35 + 273 K = 308 K

U = internal energy

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U = 2.5 n RT

U = 2.5 (2) (8.314) (308

U = 1.28 × 10⁴ J


5.

F₁ = 3300 N

F₂ = ?

A₁ = 0.060 m²

A₂ = 0.18 m²

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F₁/A₁ = F₂/A₂

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6.

R = resistance of each resistance in series = 10 ohm

R' = equivalent resistance in series

equivalent resistance is given as

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7.

R = resistance of each resistance in parallel = 10 ohm

R'' = equivalent resistance in parallel

equivalent resistance is given as

R'' = R/3

R'' = 10/3

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8.

for series circuit

R' = equivalent resistance in series = 30 ohm

V = Voltage applied = 60 Volts

i' = current in each resistor in series

current in each resistor in series is given as

I' = V/R'

i' = 60/30

i' = 2 A


for parallel circuit :

i = current in each resistor = V/R = 60/10 = 6 A








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