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# ENGINEERING APPLICATION. You designed a home that contains 1.00 x 10^5 kg of concrete [specific heat = 1.00 kJ/(kg-K)]. ...

ENGINEERING APPLICATION. You designed a home that contains 1.00 x 10^5 kg of concrete [specific heat = 1.00 kJ/(kg-K)]. How much heat is released by the concrete at night when it cools from 25.0°C to 20.0°C? 5.0 x 10^5 J. 2.0 x 10^5 kJ. 2.5 x 10^5 kJ. 5.0 x 10^5 kJ. 5.0 x 10^8 kJ. Transcribed: ENGINEERING APPLICATION. You designed a home that contains 1.00 x 10^5 kg of concrete [specific heat = 1.00 kJ/(kg-K)]. How much heat is released by the concrete at night when it cools from 25.0°C to 20.0°C? 5.0 x 10^5 J. 2.0 x 10^5 kJ. 2.5 x 10^5 kJ. 5.0 x 10^5 kJ. 5.0 x 10^8 kJ.

The heat released released or absorbed when the temperature of an material changes by ∆T is given by,

Q = ms∆T

m is mass of the material

s is special heat of the material

∆T is change in temperature

Mass of the concrete, m = 1.00 × 10^5 kg

Specific heat, s = 1.00 kJ/(kg•K)

The heat released when the concrete is cooled from T1 = 25.0°C = (273.15 + 25.0) = 298.15 K to T2 = 20.0°C = (273.15 + 20.0) = 293.15 K is

Q = ms∆T

= ms(T1 - T2)

= (1.00 × 10^5 kg)(1.00 kJ/(kg•K))(298.15 K - 293.15 K)

= (1.00 × 10^5 kg)(1.00 kJ/(kg•K))(5 K)

=  5.0 × 10^5 kJ

5.0 × 10^5 kJ

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