Q1.what is irragation
2 answers:
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Answer:
<em>a)</em> <em>1.392 x 10^6 g/cm^3</em>
<em>b) 8.69 x 10^7 lb/ft^3</em>
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Explanation:
mass of the star m = 2.0 x 10^36 kg
radius of the star (assumed to be spherical) r = 7.0 x 10^5 km = 7.0 x 10^8 m
The density of substance ρ = mass/volume
The volume of the star = volume of a sphere =
==> V = = 1.437 x 10^27 m^3
density of the star ρ = (2.0 x 10^36)/(1.437 x 10^27) = 1.392 x 10^9 kg/m^3
in g/cm^3 = (1.392 x 10^9)/1000 = <em>1.392 x 10^6 g/cm^3</em>
in lb/ft^3 = (1.392 x 10^9)/16.018 = <em>8.69 x 10^7 lb/ft^3</em>
The change in temperature (ΔT) : 56.14 ° C
<h3>Further explanation</h3>Given
Cereal energy = 235,000 J
mass of water = 1000 g
Required
the change in temperature (ΔT)
Solution
Heat can be formulated :
Q = m . c . ΔT
c = specific heat for water = 4.186 J / gram ° C
235000 = 1000 . 4.186 . ΔT