Question

A cube 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?

Answer 1

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

Step-by-step 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.

`Heat_{\text{absorbed}}=Heat_{\text{released}}`

The equation used to calculate heat released or absorbed follows:

`Q=m* c* \Delta T=m* c* (T_(final)-T_(initial))`

`m_1* c_1* (T_(final)-T_1)=-[m_2* c_2* (T_(final)-T_2)]` ......(1)

where,

q = heat absorbed or released

`m_1` = mass of ice = 12.5 g

`m_2` = mass of water = 85.0 g

`T_(final)` = final temperature = 22.24°C

`T_1` = initial temperature of ice = -15.00°C

`T_2` = initial temperature of water = 25.00°C

`c_1` = specific heat of ice = ?

`c_2` = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

`12.5* c_1* (22.24-(-15))=-[85.0* 4.186* (22.24-25)]`

`c_1=2.11J/g^oC`

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