Question

A0.350 kg iron horseshoe that is initially at 600°C is dropped into a bucket containing 21.9 kg of water at 21.8°C. What is the final equilibrium temperature (in °C)? Neglect any heat transfer to or from the surroundings. Do not enter units.

Answer 1

Answer: $22.8^0C$

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 iron horseshoe = 0.35 kg = 350 g (1kg=1000g[/tex]

$m_2$ = mass of water = 21.9 kg = 21900 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of iron horseshoe = $600^oC$

$T_2$ = temperature of water = $21.8^oC$

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

$c_2$ = specific heat of water =  $4.184J/g^0C$

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

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

$350\times 0.450\times (T_{final}-600)^0C=-[21900g\times 4.184\times (T_{final}-21.8)]$

$T_{final}=22.8^0C$

Therefore, the final equilibrium temperature is $22.8^0C$.