Question

A 360-g metal container, insulated on the outside, holds 180.0 g of water in thermal equilibrium at 22.0°C. A 24.0-g ice cube, at the melting point, is dropped into the water, and when thermal equilibrium is reached the temperature is 15.0°C. Assume there is no heat exchange with the surroundings. For water, the specific heat capacity is 4190 J/kg ∙ K and the heat of fusion is 3.34 × 105 J/kg. What is the specific heat capacity of the metal of the container?

Answer 1

Answer: specific heat capacity of the metal of the container

Cm=1143.1J/kg ∙ K

Step-by-step explanation:

Step one

Given the following data

Mass of metal container Mm=360g

Mass of water Mw=180g

Mass of ice Mice=24g

T1=22°C

T2=15°C

specific heat capacity of water

Cw = 4190 J/kg ∙ K

the heat of fusion of ice

Lf= 3.34 × 10^5 J/kg.

specific heat capacity of the metal of the container Cm=?

Step two

Heat loss by ice cube =

heat gained by water + heat gained by metal container

I.e

Mice*Lice= MwCw(T1-T2)w

+MmCm(T1-T2)

Substituting our data we have

24*3.4*10^5=180*4190*(22-15)

+360*Cm*(22-15)

8160000=754200*7+2520Cm

8160000=5279400+2520Cm

8160000-5279400 =2520Cm

2880600=2520Cm

Divide both side by 2520

2880600/2520=Cm

Cm=1143.1J/kg ∙ K