Question

Some velomobile seats have been designed to hold ice packs inside their cushions. If you started a ride with ice packs that held 1100 gg of frozen water at 0 ∘C∘C , and the temperature of the water at the end of the ride was 32 ∘C∘C , how many calories of heat energy were absorbed?

Answer 1

Answer : The heat energy absorbed will be, $1.23\times 10^5cal$

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(32.0^oC)$

The expression used will be:

$\Delta H=m\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]$

where,

m = mass of ice = 1100 g

$c_{p,l}$ = specific heat of liquid water = $4.18J/g^oC$

$\Delta H_{fusion}$ = enthalpy change for fusion = $6.01kJ/mole=6010J/mole=\frac{6010J/mole}{18g/mole}J/g=333.89J/g$

Molar mass of water = 18 g/mole

Now put all the given values in the above expression, we get:

$\Delta H=1100g\times 333.89J/g+[1100g\times 4.18J/g^oC\times (32.0-0)^oC]$

$\Delta H=514415J=122948.1358cal=1.23\times 10^5cal$

Conversion used : (1 cal = 4.184 J)

Therefore, the heat energy absorbed will be, $1.23\times 10^5cal$