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

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:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$       ......(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:

$12.5\times c_1\times (22.24-(-15))=-[85.0\times 4.186\times (22.24-25)]$

$c_1=2.11J/g^oC$

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