Question

A jar of tea is placed in sunlight until it reaches an equilibrium temperature of 30.7 ◦C . In an attempt to cool the liquid, which has a mass of 187 g , 130 g of ice at 0.0 ◦C is added. At the time at which the temperature of the tea is 24.6 ◦C , find the mass of the remaining ice in the jar. The specific heat of water is 4186 J/kg · ◦ C . Assume the specific heat capacity of the tea to be that of pure liquid water. Answer in units of g. 

Answer 1

Solution.

$m_t=187g=0.187kg;$

$t_t=30.7^oC;$

$m_i=130g=0.130kg;$

$t_i=0^oC;$

$t=24.6^oC;$

$c=4186J/kg^oC;$

$\lambda=332000J/kg;$

$c_tm_t(t_t-t)=\lambda m+c_wm(t-t_i);\implies m=\dfrac{c_tm_t(t_t-t)}{\lambda +c_w(t-t_i)};$

$m=\dfrac{4186J/kg^oC\sdot0.187kg\sdot(30.7^oC-24.6^oC)}{332000J/kg+4186J/kg^oC\sdot(24.6^oC-0^oC)}=0.011kg=11g;$

Answer: $m=11g.$