Question

Your 300 mL cup of coffee is too hot to drink when served at 90°C. What is the mass of an ice cube, taken from a -20°C freezer, that will cool your coffee to a pleasant 60°C?

Short answer

The mass is 60.2g

Step-by-step solution

Step  1: Definition of specific heat 

The amount of heat that increase1 the degree of the temperature of a 1-mole substance is called specific heat.

 Finding the amount of mass

Th equation of heat is $Q=m(c_i∆T_i+L_i+c_w∆T_w)$

95% heat has been increased in teh coffee same specific heat of the water and temperature difference $$\begin{gathered} Q=m_c(c_w∆T_c) \\ {m}_c(c_w∆T_c)=m(c_i∆T_i+L_i+c_w∆T_w) \\ m=\frac{m_c(c_w∆T_c)}{m(c_i∆T_i+L_i+c_w∆T_w)} \\ {m}_c={\rho }_{W}Vc=0.3Kg \end{gathered}$$

Applying all values

$m=\frac{4200\times 0.3\times 30}{2090\times 20+33400+4200\times 60)}=0.0602Kg$