Question

39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of the two become equal. (a) What is the final temperature (in °C) when the two become equal? (Assume that coffee has the same specific heat as water, 4.1801 J/g·°C.) °C

Answer 1

Answer: The final temperature of the solution is
`80.14^oC`

Step-by-step explanation:

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus,
`\text{heat}_(absorbed)=\text{heat}_(released)`

To calculate the amount of heat released or absorbed, we use the equation:

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

Also,

`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 aluminium = 39 g

`m_2` = mass of coffee = 166 g

`T_(final)` = final temperature = ?

`T_1` = temperature of aluminium =
`24^oC`

`T_2` = temperature of coffee =
`83^oC`

`c_1` = specific heat of aluminium =
`0.904J/g^oC`

`c_2` = specific heat of coffee=
`4.1801J/g^oC`

Putting all the values in equation 1, we get:

`39* 0.904* (T_(final)-24)=-[166* 4.1801* (T_(final)-83)]`

`T_(final)=80.14^oC`

Hence, the final temperature of the solution is
`80.14^oC`