Question

a) One way to cool a cup of coffee would be to plunge an ice-cold piece of aluminum into it. Suppose a 20.00 g piece of aluminum is stored in the refrigerator at 0.0oC and then dropped into a cup of coffee. The coffee’s temperature drops from 90.0oC to 75oC. How many kJ of thermal energy did the piece of aluminum absorb? (Specific heatAl = 0.902 J/g·oC.) 3b. Using information from question #2 above, calculate the mass (g) of coffee in the cup. (Specific heatcoffee = specific heatwater = 4.18 J/g·oC.)

Answer 1

Sol :-

(a). Calculation of heat absorbed by Al :-

Given mass of Al (m) = 20.0 g

Specific heat capacity of Al (C_s) = 0.902 J/g.⁰C

Initial temperautre of Al (T₁) = 0.0 ⁰C

Final temperature (T₂) = 75 ⁰C

Change in temperature = Δ T = T₂ - T₁ = 75 ⁰C - 0.0 ⁰C = 75⁰C

We know that

Heat absorbed by aluminium (Q) = m.C_p. Δ T = (20.0g) ( 0.902 J/g.⁰C) (75 ⁰C) = 1353 J = 1.353 KJ

Hence Heat absorbed by aluminium (Q) = 1.353 KJ

(a). Calculation of mass of coffee :-

We know

Heat lost by coffee = Heat absorbed by aluminium (Q)

So

Heat lost by coffee (Q) = m.C_p. Δ T

- 1353 J = m. (4.18 J/g·^oC) (75 - 90) ⁰C

- 1353 J = m. (4.18 J/g·^oC) (-15) ⁰C

m = - 1353 J / (4.18 J/g·^oC) (-15) ⁰C

m = 21.58 g