Question

A 94.1 g metal sample is heated to 102.4°C and transferred to 100.0 g H20 at 22.0°C in a calorimeter. Equilibrium temperature is 26.4°C. Specific heat of metal is? {SH20=4.184 J/gºC; q=smAT; AT=Tfinal-Tinitial} Answer to 2 significant figures. Do not include the unit 18°C

Answer 1

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Given m = mass of metal = 94.19 m = mass of water = 100.0g I = initial temperature of metal = 102.4°C. T2 = initial temperature of tapter = 22.0c T: Equilibrium temberachute ok wwer = 26.4°C s, = specific heat of metal = ? S2 = specific heat of water = 4.18 4 Jlg c. Since, temperature of metal is decreasting chence - metal will lose heat and temperature of water is increasing hence water will gain heat. Heat east by metal, H, " = m, s, ( T T ) 4,- 94.14.9, X 1102.4-26,4) orn, = 94. lxs, x 765 - J 0 Heat gain by water, H₂ H2= mzxSx (T-T) .. 12 100.0 X 4.184X (26.4-22) J or H, 100.0 X 2.184 x 4.4 heat gain = neat loss 94.1 xs, x 76 = 100.0 x 4.184 x 4.4 apon simplification S, .257J 0.265 sha T| ° x 1.11 Tac Scanned with Carnscanner