Question

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

Answer 1

Answer:

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.

Answer 2

Answer:

2,092 J is the amount of heat energy released when 50.0 grams of water is cooled.

Explanation:

Mass of water = m = 50.0 g

Initial temperature of the water = $T_1=20^oC$

Final temperature of the coffee = $T_2=10^oC$

Specific heat of water = c = 4.184 J/g°C

Heat required to cooled from 20°C to 10°C= Q

$Q=m\times c\times\Delta T$

$Q=mc\times (T_2-T_1)$

$Q=50.0 g\times 4.184J/g^oC\times (10^oC-20^oC)$

Q = -2,092 J

Negative sign means that energy is lost.

2,092 J is the amount of heat energy released when 50.0 grams of water is cooled.