Question

URGENT

4 pts Question 6 A 600 g piece of metal was heated to 100 °C and then put it into 100 mL of water initially at 23.7 °C). The metal and water were allowed to come to an equilibrium temperature, determined to be 278°C. Assuming no heat lost to the environment calculate the specific heat of the metal (the specific heat of water is 1.00 cal/g °C). X DE E

Answer 1

metal = malog Solution mass of metal Initial Temperature of metal = Ti = 100°C of water = v= loome initial Temperature of water = Ta = 23.7°C Eequilibrium temperature T= 27.8°c Volume = we know that, deusity of water = f = 1g/m3 so mass of water = m₂ = loog [asm=fu g We know that Heat Lost Heat gained by by metal water = mici sim m, c, AT mie, (100-27.8} = mz C2 [27-8-23-7) Gox C, x 72.2 loox lx [401] = Ci = 100 x 4.1 Gox 72.2 G = 0.095 Cellgoc so specific heat of metal is 0.095 Cal call go