Question

When 161.0 mL of water at 26.0°C is mixed with 41.0 mL of water at 85.0°C, what is the final temperature

Answer 1

Answer: The final temperature is
`38.0^0C`

Step-by-step explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

`heat_(released)=heat_(absorbed)`

`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)]`

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

Q = heat absorbed or released

`m_1` = mass of water at
`85.0^0C` =
`volume* density=41.0ml* 1g/ml=41.0g`

`m_2` = mass of water at
`26.0^0C` =
`volume* density=161.0ml* 1g/ml=161.0g`

`T_(final)` = final temperature = ?

`T_1` = temperature of 41.0 ml of water =
`85.0^0C`

`T_2` = temperature of 161.0 ml of water =
`26.0^0C`

Now put all the given values, we get

`-[41.0* (T_f-85.0)^0C]=161.0* (T_f-26.0)^0C`

`T_f=38.0^0C`

Thus the final temperature is
`38.0^0C`