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STALIN [3.7K]
3 years ago
10

20950 joules of heat are added to a container with 400 g of water. how much does the temperature change??? Show work and formula

.
Chemistry
1 answer:
krok68 [10]3 years ago
7 0

Answer:

The temperature change is 12.5 °C

Explanation:

Step 1: Data given

Energy added to the water = 20950 joules of heat

Mass of water = 400 grams

The specific heat of water = 4.184 J/g°C

Step 2: Calculate the change in temperature

q = m*c*ΔT

⇒ with q = the energy (heat) added to the container = 20950 J

⇒ with m = the mass of water = 400 grams

⇒ with c = the specific heat of water = 4.184 J/g°C

⇒ with ΔT = the change of temperature = TO BE DETERMINED

20950 J = 400 * 4.184 * ΔT

ΔT = 20950 J / (400*4.184)

ΔT = 12.5 °C

The temperature change is 12.5 °C

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If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i
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Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

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Butane  ⇄  Isobutane

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1.0 M               2.5 M

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(1.50-x)M            (2.5+x)

The equilibrium expression will wriiten as:

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