Question

what is the final temperatyre of both the weight and the water at thermal equilibrium

MHLETULLI A2.09 g lead weight, initially at 10.3 °C. is submerged in 8.04 g of water at 52.3°C in an insulated container. You may want to reference (Page) Section 6.4 while completing this problem.

specific heat is 4.814 J

Answer 1

LCW) Thermal final temperature of both the weight and water at thermal equilibruim. (Tf) By Given information: Lead weight = 2.099 at 10.3°C. 8.04g water in 52.3°C. Specific heat of water 4.1845/g °C specific heat of lead. (cpb) = 0.128 J/g °c. By conservation of energy Apb & qw=0 qa heat energy & Qpb = -Qw [9p6 = mpb Cpg STP] m represent mass eng [ {w = mw O w OTw] LT is change in temperature in oc. flere cold metal cool down the solution by absorbing heat from the water we know mw-roug Co= 4.1 845/9°c. Tp6 = 10.3°C Tw= 52.3°C CamScanner and Cpb = Q. 128 J/g °c mpb= 2.099

The equation becomes Mpb C Pb (Tf-Top)=– mw cw IT, - Tw) on sustituting values (209), (0.128) 5/8°C (tp - 10.3°) = - 8.04 g (4.184) J19c (Tf - 52.3°C) Omitting units for simplicity a) 0.267 ( If - 10.3) = - 33.639 (If -52-3) , 0.267 T6 - 10.310.261) = -33.639 76 + 152-3)(33639) 2) 0.267 T4 - 2.750 = -33.639 TK + 1759.319 20.267 T6 +33 639 Tb = 1759.319 +2.750 33.90 14 - 1762.06 » Tb = 1762.06 = 51.9°C 33.90 head has such a that i t cannot energy as water cs x 152.0 Cappoxo) low heat of capacity store as much thermal can. Scan Cams