Question

A 100.0 g ice cube at -10 degrees Celsius is placed in an aluminum cup whose initial temperature is 70 degrees Celsius. The system come to an equilibrium at 20 degrees Celsius. What is the mass of the cup?

Answer 1

Answer: 135 grams

Explanation:

$Q_{absorbed}=Q_{released}$

As we know that,  

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$    

where,

$m_1$ = mass of ice = 100 g

$m_2$ = mass of aluminium cup =? g

$T_{final}$ = final temperature  =$20^0C$

$T_1$ = temperature of ice = $-10^oC$

$T_2$ = temperature of aluminium cup= $70^oC$

$c_1$ = specific heat of ice= $2.03J/g^0C$

$c_2$ = specific heat of aluminium cup = $0.902 J/g^0C$

Now put all the given values in equation (1), we get

$[100\times 2.03\times (20-(-10))]=-[m_2\times 0.902\times (20-70)]$

$m_2=135g$

Therefore, the mass of the aluminium cup was 135 g.