Question

179.1 g of water is in a Styrofoam calorimeter of negligible heat capacity. The initial T of the water is 16.1oC. After 306.9 g of an unknown compound at 94.3oC is added, the equilibrium T is 27.4oC.

What is the specific heat of the unknown compound in J/(goC)?

Answer 1

the specific heat of the unknown compound is
`c_u=0.412J/g \cdot C`

Step-by-step explanation:

Generally the change in temperature of water is evaluated as

`\Delta T = T_2 -T_1`

Substituting 16.1°C for
`T_1` and 27.4°C for
`T_2`

`\Delta T = 27.4 - 16.1`

`=11.3^oC`

Generally the change in temperature of unknown compound is evaluated as

`\Delta T_u = T_3 -T_2`

Substituting 27.4°C for
`T_2` and 94.3°C for
`T_3`

`\Delta T = 94.3 - 27.4`

`=66.9^oC`

Since there is an increase in temperature then heat is gained by water and this can be evaluated as

`H_w = mc_w \Delta T`

Substituting 179.1 g for m , 4.18 J/g.C for
`c_w`(specific heat of water)

`H_w = 4.18 * 179.1 * 11.3`

`= 8459.6J`

Since there is a decrease in temperature then heat is lost by unknown compound and this can be evaluated as

`H_u = m_uc_u \Delta T_u`

By conservation of energy law

Heat lost = Heat gained

Substituting 306.9 g for
`m_u` , 8459.6J for
`H_u`

`8459.6 = 306.9 * c_u * 66.9`

Therefore
`c_u = (8459.6)/(308.9 *66.9)`

`=0.412J/g \cdot C`