Question

2. A sample of lead is heated to 189.4°C. It is then placed in a cup containing 32.8 g water at 24.3°C. The final temperature of the water and lead is 29.6°C Calculate the mass of the sample of lead. The specific heat of water is 4.184 Jig C and the specific heat of lead is 0.130 J/g .

Answer 1

Thanks, please upvote

> Solutim:- to let mass of sample of lead is 'migram. Given dlater Initial Temperature of heard sample = 189.4°C final Temperature of water and lead = 29.6 °e Therefore, change in Temperature of leno) sample— AT,= YRAMY (2916–189.4)ė = -159.8°C specific heat of lead sample, ep.= 0.130 T i Mass of water in cup, m2 = 32.8 gram.! specific heat of teater, CP2 = 4.184 I Initial Temperature of water. -= 24.3%ce - change in Temp. of water, Ate = €9.6-243) e >> Now: we observed that, Temp. for heart of Metal decreases AT2 = 5.3 °C while temp/Heat of water increases. Hence, Atequilibrium - & metal = . Vwater are sign for decrease. om -mep, ot, a.m2 CPLATE = Mixology at 159.85$ = 32-8gx 4.1845.36 = +20.774 x mi = 727.346 am Blog = 3siong of lead sample >>) me = 724346