Question

A 452.3-g sample of an element at 192°C is dropped into an ice–water mixture; 110.6 g of ice melts and an ice–water mixture remains. Calculate the specific heat of the element. ΔHfusion = 6.02 kJ/mol (for liquid water at 0°C).

Answer 1

the specific heat of the sample is c sam = 0.42 J/g °C

Step-by-step explanation:

if we assume that the sample is submerged into the mixture, such that all the heat released by the sample is absorbed by the ice and water, and there is no time to the heat to be released to the surroundings ( or the system is isolated):

Q ice wat + Q sam = Q surroundings = 0

Q ice wat= - Q sam

since the ice-water mixture remains, the final temperature is 0°C (water-ice equilibrium), thus only latent heat is involved

Assuming that there is no heat due to reaction or phase change of the sample, the heat released is only sensible heat

Q sam = m sam * c sam ( T final - T initial )

Q ice wat = m ice * ΔH fusion / M water

therefore

- m sam * c sam ( T final - T initial ) = m ice * ΔH fusion / M water

c sam= m ice * ΔH fusion /[ m sam * M water *( T initial - T final )]

replacing values

c sam = 110.6 g * 6.02 kJ/mol / ( 452.3 g * 18 gr/mol*( 192°C -0°C) * 1000 J/kJ = 0.42 J/g °C

thus

c sam = 0.42 J/g °C