Question

A Lead-tin alloy, 150g at 100 degrees celsius is placed in a 50.0g H2O at 22 degrees celsius in an insulated flask. When thermal equilibrium is reached, T=28.8 degrees celsius. The specific heat capacity of water is 4.18J/g k. What is the specific heat capacity of the alloy?

Answer 1

This is an example of heat balance problem.

Heat lost by alloy = heat gained by water, we can use equation

mcΔT = Q: m = mass; C_s = specific heat capacity: ΔT = temp. difference; Q = Heat absorbed or Released.

m_alloy x C_s(alloy) x (T_alloy - T_final) = m_water x C_s (water) x (T_final - T_water)

150 g x C_s(alloy) x (100 - 28.8) = 50 g x 4.18 J/g.K x (28.8 - 22.0)

C_s(alloy) = [50 g x 4.18 J/g.K x (28.8 - 22.0)] / (150 g x (100 - 28.8))

C_s(alloy) = 0.13307116104 J/g.K

specific heat capacity of the alloy (C_s(alloy)) = 0.1331 J/g.K

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