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* c* \Delta T=m* c* (T_(final)-T_(initial))`

`m_1* c* (T_(final)-T_1)=-[m_2* c* (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* c_1* (T_(final)-T_1)=-[m_2* c_2* (T_(final)-T_2)]`

`350* 0.450* (T_(final)-600)^0C=-[21900g* 4.184* (T_(final)-21.8)]`

`T_(final)=22.8^0C`

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