Question

A sample of 14.0 g manganese, originally at 80 degrees C, is dropped into 49.0 g of water at 20 degrees C. The molar heat capacity of Mn is 29.4 J/mol*C. What is the final temperature of the water?

Answer 1

molar heat capacity of Mn = 29.4 J/mol.oC

molar mass of Mn = 54.94 g/mol

use:

specific heat capacity of Mn = molar heat capacity of Mn / molar mass

= (29.4 J/mol.oC)/(54.94 g/mol)

= 0.535 J/g.oC

Let us denote water by symbol 1 and Manganese by symbol 2

m1 = 49.0 g

T1 = 20.0 oC

C1 = 4.184 J/goC

m2 = 14.0 g

T2 = 80.0 oC

C2 = 0.535 J/goC

T = to be calculated

Let the final temperature be T oC

we have below equation to be used:

heat lost by 2 = heat gained by 1

m2*C2*(T2-T) = m1*C1*(T-T1)

14.0*0.535*(80.0-T) = 49.0*4.184*(T-20.0)

7.49*(80.0-T) = 205.016*(T-20.0)

599.2 - 7.49*T = 205.016*T - 4100.32

T= 22.1 oC

Answer: 22.1 oC