In a glass aluminum vessel with a mass of m1 = 2 kg, and having a temperature of t1 = 40 ° C, m2 = 3 kg

| April 28, 2021 | education

In a glass aluminum vessel with a mass of m1 = 2 kg, and having a temperature of t1 = 40 ° C, m2 = 3 kg of water was poured with a temperature of t2 = 15 ° C. The vessel with water was placed in the sun. After τ = 2 hours its temperature became t = 35 ° С. The power of solar radiation falling on the vessel is 200 watts. Find what fraction of the solar energy it absorbs. Disregard the interaction of the vessel with the environment.

Change in the internal energy of the vessel: qa = Ca * m1 (t-t1) where Ca is the heat capacity of aluminum (Ca = 920 J / kg * deg), and for water qw = Cw * m2 (t-t2), Cw is the heat capacity
water Cw = 4200 J / kg * deg.

The total absorbed energy is:
qa + qw = Ca * m1 (t-t1) + Cw * m2 (t-t2) = 920 * 2 * (35-40) + 4200 * 3 (35-15) = – 9200 + 25200 = 242800J

The total amount of solar energy during this time: 200 * 2 * 3600 = 1440000.

Then we get:
242800/1440000 = 0.167



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