Question

A hot lump of 32.3 g of copper at an initial temperature of 96.5°C is placed in 50.0 mL H2O initially at 25.0°C and allowed to reach thermal equilibrium. What is the final temperature of the copper and water given that the specific heat of copper is 0.385J/g°C and the specific heat of water is 4.184J/g°C?

4. A hot lump of 32.3 g of copper at an initial temperature of 96.5°C is placed in 50.0 mL H20 initially at 25.0°C and allowed to reach thermal equilibrium. What is the final temperature of the copper and water, given that the specific heat of copper is 0.385 J/g°C and the specific heat of water is 4.184 J/g°C? a. 24.5°C b. 26.0°C c. 27.5°C d. 29.0°C S-80

Answer 1

= 50 ML mass of Cu = mau= 32.3 g. Initial temperature of a = tem = 96.5 c specific heat of Cu = 5 = 0.385 J/g °c Again, volume of water = 50 min i mass of water = m. = 50 ml x 1 9/m2 = 50g density of . [water = 1g/ml Initial Temperature of water Ho = 25°c Specific heat of water, Shoo - 4.184 dl goc Let the final temperature of the system be – toc

Heat lost by cu A = - Cu = - Men Seult - 96.5) heat gained by water = + q to = + mai Stogo (t - 25) Acconding to the principle of Calo- roimefroy - Musea (+-96-5)= tm Star (+-25) a - 32.3 g x 0.385 5. 19o. x(+-965) = 50 f x 4.184 D9%x (+-25) a - 12.4 t + 1200 = 209.2t - 5230 a 209. 2t +12.9+ = 1200+5230 a 221.6 t = 6430 a t - 64% = 29.03. 221.6

Thus the final temperature of the copper and water be - d. 29.0°c) Ais