Question

A 26.0-g aluminum block is warmed to 65.1 ∘C and plunged into an insulated beaker containing 55.5 g of water initially at 22.3 ∘C . The aluminum and the water are allowed to come to thermal equilibrium. (Cs,H2O=4.18J/g⋅∘C , Cs,Al=0.903J/g⋅∘C )

Assuming that no heat is lost, what is the final temperature of the water and aluminum?

T = ??      ∘C     

Answer 1

Heat =Q = m c (delta T)

Mass of Aluminum block = 26.0 g

Specific heat of Aluminum block = 0.903 J/g⁰C

Initial temperature of copper = 65.1 ⁰C

Final temperature of copper = 22.3 ⁰C

So, Heat lost by copper , Q_Cu = mc (delta T) = (26.0 g) x (0.903 J/g⁰C) x (22.3 - 65.1) ⁰C)

                                                            = - 1004.86 J

Since this is the heat lost, so its value will be negative of the value calculated.

So, - (- 1004.86 J) = 1004.86 J

Mass of Water = 55.5 g

Specific heat of water is 4.18 J/g⁰C

Initial temperature of water = 22.3 ⁰C

Final temperature of water = T ⁰C

So, Heat gained by waterr , Q_H2O = mc (delta T) = (55.5 g) x (4.18 J/g⁰C) x (T - 22.3) ⁰C)

                                                                        = 231.99 x (T - 22.3) J

Since, Heat lost = Heat gained

Q_Cu = Q_H2O

1004.86 J = 231.99 x (T - 22.3) J

(T - 22.3) = 1004.86 /(231.99) = 4.33

T = 4.33 + 22.3 = 26.63

Hence, the final temperature, T is 26.63 ⁰C