Question

A 32.4 g iron rod, initially at 21.6 ∘C, is submerged into an unknown mass of water at 63.1 ∘C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 59.6 ∘C. You may want tWhat is the mass of the water?

Answer 1

Answer:  37.8 grams

Explanation:-

$heat_{absorbed}=heat_{released}$

As we know that

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

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$         .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of iron rod = 32.4 g

$m_2$ = mass of water = ?

$T_{final}$ = final temperature =  $59.6^oC=332.6K$

$T_1$ = temperature of iron rod = $21.6^oC=294.6K$

$T_2$ = temperature of water = $63.1^oC=336.1K$

$c_1$ = specific heat of iron rod = $0.450J/g^0C$

$c_2$ = specific heat of water= $4.18J/g^0C$

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

$32.4\times 0.450\times (332.6-294.6)=-[m_2\times 4.18\times (332.6-336.1)]$

$m_2=37.8g$

Therefore, the mass of the water is 37.8 grams