table.jadata{border-bottom: 1px solid black; border-top: 1pxsolid black;} td.jad

Question

table.jadata{border-bottom: 1px solid black; border-top: 1pxsolid black;} td.jadata{padding-left: 8px; padding-right: 8px;}th.jadata_b{padding-left: 8px; padding-right: 8px; border-bottom:1px solid black;} td.jadata_r{border-right: 1px solid black;}td.jadata_brl{border-bottom: 1px solid black; padding-right: 8px;border-right: 1px solid black; border-left: 1px solid black;}th.jadata_br{border-bottom: 1px solid black; padding-right: 8px;border-right: 1px solid black;}

Hfus(0°C)(H2O, s) 6008 J/mol CP,m(H2O, l) 75.3 J · K 1 ·mol 1 CP,m(H2O, s) 36.8 J · K 1 ·mol 1

Explanation / Answer

                According tothe law of conservation of energy the heat lost by the warmsubstance is equal to the heat gained by the cold substance .                100g *CP,m(H2O, s) * ( T -(-15oC) = 200g *CP,m(H2O, l) *(25oC- T)       ( 100g /18g/mol) *36.8  J · K 1 ·mol 1 * ( T - (-15oC) = (200g/18g/mol) * 75.3 J · K 1 ·mol 1 * (25-T )                        final temperature of the mixture = T = 17.144oC.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---