Question

A 10.000g ice cube is added to 100.000g of water at 80.0 oC in a calorimeter. The final temperature of the water in the calorimeter is 65.5 oC. How much heat in kilojoules was needed to melt the ice cube?

Answer 1

Answer: The amount of heat needed to melt the ice cube is 6.083 kJ

Explanation:

The processes involved in the given problem are:

$1.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\2.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(65.5^oC,338.5K)$

• For process 1:

To calculate the amount of heat required to melt the ice at its melting point, we use the equation:

$q_1=m\times \Delta H_{fusion}$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of ice = 10.000 g

$\Delta H_{fusion}$ = enthalpy change for fusion = 334.16 J/g

Putting all the values in above equation, we get:

$q_1=10.000g\times 334.16J/g=3341.6J$

• For process 2:

To calculate the heat required at different temperature, we use the equation:

$q_2=mc\Delta T$

where,

$q_2$ = heat absorbed

m = mass of ice = 10.000 g

c = specific heat capacity of water = 4.184 J/g.°C

$\Delta T$ = change in temperature = $T_2-T_1=[65.0-0]^oC=65.5^oC$

Putting values in above equation, we get:

$q_2=10.000g\times 4.184J/g.^oC\times 65.5^oC\\\\q_2=2741.83J$

Total heat absorbed = $q_1+q_2$

Total heat absorbed = $[3341.6+2741.83]J=6083.43J=6.083kJ$

Hence, the amount of heat needed to melt the ice cube is 6.083 kJ