Question

Q17

A 17.3−gsample of an alloy at 93.00°Cis placed into 50.0 gof water at 22.00°Cin an insulated coffee-cup calorimeter with a heat capacity of 9.20 J/K.If the final temperature of the system is 31.10°C,what is the specific heat capacity of the alloy?

(---------) J/g·°C

Answer 1

We know that,

Heat lost by alloy = heat gained by water

m(alloy)*C(alloy)*(T(alloy)-T) = m(water)*C(water)*(T-T(water)) + C3*(T-T(water))

Given data,

mass of (alloy) =17.3 g

Temperature of (alloy) = 93.0 oC

mass of (water) = 50.0 g

Temperature of (water) = 22.0 oC

C(water) = 4.184 J/goC

C3 = 9.2 J/oC

Substitute these values in the above equation, we get

17.3 * C(alloy) * (93.0-31.1) = 50.0 * 4.184 * (31.1-22.0) + 9.2 * (31.1-22.0)

1070.87 * C(alloy) = 1903.72 + 83.72

1070.87 * C(alloy) = 1987.44

C(alloy)= 1.855 J/goC