Question

How much energy is required to change a 39 g ice cube from ice at −14◦C to steam at 114◦C? The specific heat of ice is 2090 J/kg · ◦ C, the specific heat of water is 4186 J/kg · ◦ C, the specific heat of stream is 2010 J/kg · ◦ C, the heat of fusion is 3.33 × 105 J/kg, and the heat of vaporization is 2.26 × 106 J/kg. Answer in units of J. 

Answer 1

Solution.

$m_i=39g=0.039kg;$

$t_i=-14^oC;$

$t_s=114^oC;$

$c_i=2090J/kg^oC;$

$c_w=4186J/kg^oC;$

$c_s=2010J/kg^oC;$

$\lambda=330000J/kg;$

$r=2260000J/kg;$

$Q=Q_1+Q_2+Q_3+Q_4+Q_5;$

$Q=c_im_i\Delta t_i+\lambda m_i+c_wm_i\Delta t_w+rm_i+c_sm_i\Delta t_s;$

$Q=m_i(c_i\Delta t_i+\lambda +c_w\Delta t_w+r+c_s\Delta t_s);$

$Q=0.039kg(2090J/kg^oC\sdot14^oc+330000J/kg+4186J/kg^oC\sdot100^oC+2260000J/kg+2010J/kg^oC\sdot14^oC)=119574J;$

Answer: $Q=119574J.$