Question

A 25.0−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?

Answer 1

1.23 J/(g °C)

Step-by-step explanation:

`m_(w)` = mass of water = 50 g

`c_(w)` = specific heat of water = 4.186 J/(g °C)

`T_(wo)` = Initial temperature of water = 22.00 °C

`m_(a)` = mass of alloy = 25 g

`c_(a)` = specific heat of alloy = ?

`T_(ao)` = Initial temperature of alloy = 93.00 °C

`T_(f)` = Final equilibrium temperature = 31.10 °C

Using conservation of heat

Heat Lost by alloy = Heat gained by water

`m_(a)c_(a)(T_(ao)-T_(f))` =
`m_(a)c_(a)(T_(f) - T_(ao))`

(25)
`c_(a)` (93 - 31.10) = (50) (4.186) (31.10 - 22)

`c_(a)` = 1.23 J/(g °C)