What is the amount of heat energy released when 50.0 grams of water is cooled from 20.0 degrees celcius to 10.0 degrees celcius

The amount of heat energy released is -2092 J

From the question,

We are to determine the amount of heat energy released when 50.0 g of water is cooled from 20.0 °C to 10.0 °C

From the formula

Q = mcΔT

Where Q is the quantity of heat

m is the mass of substance

c is specific heat capacity of substance (c of water = 4.184 J/g °C)

ΔT is the change in temperature

From the given information

m = 50.0 g

ΔT = T₂ - T₁ = 10.0 °C - 20.0 °C

ΔT = -10.0 °C

Putting these values into the formula, we get

Q = 50.0 × 4.184 × -10.0

Q = -2092 J

Hence, the amount of heat energy released is -2092 J

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The amount of heat released when 50 g of water cooled from 20°C to 10°C will be equal to  - 2093 J.

Explanation:

Given data:

Mass of water = 50 g

Initial temperature= T1 = 20°C

Final temperature= T2 = 10°C

Specific heat of water= c = 4.186 J/g. °C

Amount of heat released = Q= ?

Solution:

Formula:

Q = m. C.  ΔT

ΔT = T2 - T1

ΔT = 10°C - 20°C

ΔT = -10°C

Now we will put the values in formula.

Q = m. C.  ΔT

Q = 50 g .  4.186 J/g. °C . -10°C

Q =  - 2093 J

The amount of heat released when 50 g of water cooled from 20°C to 10°C will be equal to  - 2093 J.

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