Acube of ice at an initial temperature of -15.00°c weighing 12.5 g total is placed in 85.0 g of water at an initial temperature of 25.00°c. when thermal equilibrium is achieved, the final temperature is 22.24°c. what is the specific heat capacity of ice?

The specific heat of ice is 2.11 J/g°C

Explanation:

When ice is mixed with water, the amount of heat released by water will be equal to the amount of heat absorbed by ice.

The equation used to calculate heat released or absorbed follows:

      ......(1)

where,

q = heat absorbed or released

= mass of ice = 12.5 g

= mass of water = 85.0 g

= final temperature = 22.24°C

= initial temperature of ice = -15.00°C

= initial temperature of water = 25.00°C

= specific heat of ice = ?

= specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

Hence, the specific heat of ice is 2.11 J/g°C

pe = weightxheight = 2000x1000=2,000,000joules