Exercise 7 | Thermochemistry

Water: higher temperature ⇒ transfer energy to aluminum

Aluminum: lower temperature ⇒ gain energy from water

q_lost (water) + q_gained (aluminum) = 0

c_p (water) x (T_f – T_i (water)) = - c_p (Al) x (T_f – T_i (Al))

⇒ c_p (water) x (T_f – T_i (water)) = - c_s (Al) x m (Al) x (T_f – T_i (Al))

⇒ c_s (Al) = - $\frac{{\mathrm{c}}_{\mathrm{P}} \left(\mathrm{water}\right) \times \left({\mathrm{T}}_{\mathrm{f}} - {\mathrm{T}}_{\mathrm{i}} \left(\mathrm{water}\right)\right)}{\mathrm{m} \left(\mathrm{Al}\right) \times \left({\mathrm{T}}_{\mathrm{f}} - {\mathrm{T}}_{\mathrm{i}} \left(\mathrm{Al}\right)\right)}$

⇒ c_s (Al) = - $\frac{417.9 \times (331.3 - 338)}{35.0 \times (331.3 - 298)}$

⇒ c_s (Al) = 2.40 J.g^-1.K^-1