Question

25 g of 116oC steam are bubbled into 0.2384 kg of water at 8oC. Find the final temperature of the system.

Answer 1

Answer: The final temperature of the system will be
`13.14^0C`

Step-by-step explanation:

`heat_(absorbed)=heat_(released)`

As we know that,

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

`m_1* c_1* (T_(final)-T_1)=-[m_2* c_2* (T_(final)-T_2)]` .................(1)

where,

q = heat absorbed or released

`m_1` = mass of steam = 25 g

`m_2` = mass of water = 0.2384 kg = 238.4 g (1kg=1000g)

`T_(final)` = final temperature = ?

`T_1` = temperature of steam =
`116^oC`

`T_2` = temperature of water =
`8^oC`

`c_1` = specific heat of steam =
`1.996J/g^0C`

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

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

`25g* 1.996J/g^0C* (T_(final)-116)=-[238.4g* 4.184J/g^0C* (T_(final)-8)]`

`T_(final)=13.14^0C`

Therefore, the final temperature of the system will be
`13.14^0C`