Question

28. A hot ball of steel weighing 100g is dropped into 600 g of liquid at 20 °C. The resulting temperature is
32 °C. Calculate the specific heat of liquid if the temperature of hot ball was 400 °C and specified heat
of steel is 312 J kg 'K'.
[1594.6 J kg 'K'​

Answer 1

Given:

A hot ball of steel weighing 100g is dropped into 600 g of liquid at 20 °C. The resulting temperature is 32 °C.

To find:

Specific heat of liquid.

Calculation:

Let $\theta$ be equilibrium temperature and c2 be specific heat capacity of liquid.

Applying Expression for Calorimetry:

$\rm{ \therefore \: (m1)(c1)(\Delta \theta_{1}) = (m2)(c2)(\Delta \theta_{2})}$

$\rm{ = > \: (m1)(c1)( \theta_{1} - \theta) = (m2)(c2)( \theta - \theta_{2})}$

$\rm{ = > \: (100)(312)( 400 -32) = (600)(c2)( 32 - 20)}$

$\rm{ = > \: (100)(312)( 368) = (600)(c2)( 12)}$

$\rm{ = > \: (312)( 368) =72(c2)}$

$\rm{= > c2 = 1594.67 \: J{kg}^{ - 1} {K}^{ - 1} }$

So, final answer is:

$\boxed{ \large{ \bf{ c2 = 1594.67 \: J{kg}^{ - 1} {K}^{ - 1} }}}$