Question

If we place a solid hot metal in a cool liquid the metal and the liquid come to some final temperature together. The specific heat of the liquid is five times as great as the specific heat of the metal. The temperature change for the metal is twenty times as much as that for the liquid.
How does the mass of the liquid compare to the mass of the solid metal? Assume no heat is lost from the system.

• A. The mass of the liquid is 100 times as much as the mass of the metal.
• B. The mass of the liquid is 5 times as much as the mass of the metal.
• C. The mass of the liquid is 20 times as much as the mass of the metal.
• D. The mass of the liquid is 1/4 times as much as the mass of the metal.
• E. The mass of the liquid is 4 times as much as the mass of the metal.

Answer 1

Answer: (E)

The mass of the liquid is 4 times as much as the mass of the metal.

Since there is no heat loss from the system, the heat lost by hot metal goes into the cool liquid.

Hence $m_s{s}_s\Delta T_s=m_l{s}_l\Delta T_l$

$⟹\frac{m_l}{m_s}=\frac{s_s\Delta T_s}{s_l\Delta T_l}$

$=\frac{1}{5}.\frac{20}{1}=4$