What temperature will 1L of H20 at 200°F become when a piece of copper,0.25kg at 260.928K, comes into contact with water?

Explanation:

mass of 1 L water = 1 kg .

200⁰F = (200 - 32) x 5 / 9 = 93.33⁰C .

260.928 K = 260.928 - 273 = - 12.072⁰C .

water is at higher temperature .

Let the equilibrium temperature be t .

Heat lost by water = mass x specific heat x  fall of temperature

= 1 x 4.2 x 10³ x ( 93.33 - t )

Heat gained by copper

= .25 x .385 x 10³ x ( t +  12.072 )

Heat lost = heat gained

1 x 4.2 x 10³ x ( 93.33 - t ) = .25 x .385 x 10³ x ( t +  12.072 )

93.33 - t = .0229 ( t + 12.072)

93.33 - t = .0229 t + .276

93.054 = 1.0229 t

t = 90.97⁰C .