Question

Substance gold A silver cube with an edge length of 2.28 cm and a gold cube with an edge length of 2.65 cm are both heated to 89.1 °C and placed in 118.0 mL of water at 19.7 °C. What is the final temperature of the water when thermal equilibrium is reached? Specific heat (J/g•°C) 0.1256 0.2386 4.184 Density (g/cm) 19.3 10.5 1.00 silver water Tfinal =

Answer 1

(= edge) volume of silver lube = (2.2883 cm3 volume of Gold Cube = (2.653 am3 Density of Gold a 19.3 g/cm3 mass of Gold Gibe = Density & volume = 19.3 $ (2.65) g = 359.1669 mass of selver Cube = 10.5*(2.2893g = 124145g let's say final temperature is T. (in oc) At= change in Temperature Meat lost by Gold Cube = m .sgold X AT Sgold = specific heat of = 359 X01256 (89.1-T) 5011256 = 45.11 (89.1-T) J for silver, Heat lost = 124.45 X 0.2386 1 89.1 -T) J = 29.69(89.1-7) J . Goldf 59 - Heat Gained by water = msat m= mass of water = specific heat = 1188 4.1848 (T-19.7) 3 493.712 (T-19.7) Heat Gained = Total Heat lost * 493.712 (1-19.7) = 45:11 (89.1-1) + 29.69(89.1-T) > 1T= 28.83°c)