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3 years ago

If a 375 mL sample of water was cooled from 37.5 C to 0 C how much heat wad lost in joules

Physics

2 answers:

Aleks [24]3 years ago

5 0

The equation for heat is
Q = m*Cp*del(T) where
Q = heat in Joules
m = mass in kg
Cp = the heat capacity of water at a constant pressure = 4,184 J/kgC
del(T) = the change in temperature in either K or degrees C
1 liter of water = 1 kg at standard temperature and pressure.
375 mL water x (1 kg/1L water) = 0.375 kg water.
Q = (0.375 kg water)(4184 J/kgC)(37.5 C - 0 C) = 58,800 Joules.

harkovskaia [24]3 years ago

3 0

1 mL = 1 gram
Q=375*4.179*37.5

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2. 160 N.

3. 0 N.

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1. Determination of the net force.

Force applied to the right (Fᵣ) = 80 N

Force applied to the left (Fₗ) = 145 N

Net force (Fₙ) =?

Fₙ = Fₗ – Fᵣ

Fₙ = 145 – 80

Fₙ = 65 N

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2. Determination of the net force.

Force 1 applied to the left (F₁) = 35 N

Force 2 applied to the left (F₂) = 125 N

Net force (Fₙ) =?

Fₙ = F₁ + F₂

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Thus, the net force is 160 N.

3. Determination of the net force.

Force applied to the right (Fᵣ) = 75 N

Force applied to the left (Fₗ) = 75 N

Net force (Fₙ) =?

Fₙ = Fₗ – Fᵣ

Fₙ = 75 – 75

Fₙ = 0

Thus, the net force is 0 N

4. Determination of the net force.

Force 1 applied to the right (F₁) = 150 N

Force 2 applied to the right (F₂) = 60 N

Net force (Fₙ) =?

Fₙ = F₁ + F₂

Fₙ = 150 + 60

Fₙ = 210 N

Thus, the net force is 210 N.

5. Determination of the net force.

Force applied to the right (Fᵣ) = 115 N

Force applied to the left (Fₗ) = 25 N

Net force (Fₙ) =?

Fₙ = Fᵣ – Fₗ

Fₙ = 115 – 25

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Thus, the net force is 90 N

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