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a 2.00x10^2 -g sample of water at 60 C is heated yo water vapor at 140 C. How much thermal energy is absorbed

Sagot :

Answer:

The thermal energy absorbed is 66.98 kJ.

Explanation:

The thermal energy absorbed can be found by using the following equation:

[tex] Q = mC\Delta T [/tex]

Where:

Q: is the thermal energy =?

m: is the mass of water = 2.00x10² g

ΔT = T₂ - T₁ = 140 °C - 60 °C = 80 °C

C: is the specific heat of water = 4.186 J/(°C*g)

Hence, the thermal energy absorbed is:  

[tex]Q = mC\Delta T = 2.00 \cdot 10^{2} g*4.186 \frac{J}{^{\circ} C*g}*80 ^{\circ} C = 66976 J = 66.98 kJ[/tex]

Therefore, the thermal energy absorbed is 66.98 kJ.

I hope it helps you!    

The thermal energy absorbed will be 66.98 kJ.heat loss is inverse to heat gain.

What are heat gain and heat loss?

Heat gain is defined as the amount of heat required to increase the temperature of a substance by some degree of Celcius.

Given data;

Q(thermal energy) =?

m(mass of water) = 2.00x10² g

C(specific heat of water) = 4.186 J/(°C*g)

ΔT = T₂ - T₁

ΔT = 140 °C - 60 °C

ΔT = 80 °C

It is given by the formula as ;

[tex]\rm Q= mcdt \\\\ Q= 2.00x10^2 \times 4.186 \times 80 \\\\ Q=66.98 \ kJ.[/tex]

Hence, the thermal energy absorbed will be 66.98 kJ.

To learn more about the heat gain refer to the link;

https://brainly.com/question/26268921

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