Question

e 6.1 kg of water at ooc absorbs of heat to convert into steam at 100°C​

Answer 1

Answer:

From first law of thermodynamic ΔQ=ΔU+W

ΔQ=mL=1×540cal.=540cal.

W=PΔV=

4.2

10

6

(1671−1)×10

−6

=

4.2

10

5

×(1670)×10

−6

≈40cal.

ΔU=540−40=500cal.

Answer 2

Answer:

Given

Mass - 6.1 Kg

Temp - 100 c

To find Heat required.

Q = m * L water .

Here L Water - Heat of vaporization.

When a material in liquid state is given energy, it changes its phase from liquid to vapor; the energy absorbed in this process is called heat of vaporization. The heat of vaporization of water is about 2,260 kJ/kg.

Q = 1 * 2260

Q = 2260 kJ/kg

Given

Mass - 6.1 Kg

Temp - 0 c

To find Heat required.

Q = m * L ice.

Here L Water - Latent heat of ice .

while ice melts, it remains at 0 °C (32 °F), and the liquid water that is formed with the latent heat of fusion is also at 0 °C. The heat of fusion for water at 0 °C is approximately 334 joules

Q = 1 * 334

Q = 334 kJ/kg

Thanks

Explanation:

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