Question

A chunk of tin weighing 18.5 grams and originally at 97.38 °C is dropped into an insulated cup containing 75.7 grams of water at 21.52 °C. Assuming that all of the heat is transferred to the water, the final temperature of the water is ____

Answer 1

The final temperature of the water is 22.44°C.

Step-by-step explanation:

Heat lost by tin will be equal to heat gained by the water

`-q_1=q_2`

Mass of tin =
`m_1=18.5 g`

Specific heat capacity of tin =
`c_1=0.21 J/g^oC`

Initial temperature of the tin =
`T_1=97.38^oC`

Final temperature =
`T_2`=T

`q_1=m_1c_1* (T-T_1)`

Mass of water=
`m_2=75.7 g`

Specific heat capacity of water=
`c_2=4.184 J/g^oC`

Initial temperature of the water =
`T_3=21.52^oC`

Final temperature of water =
`T_2`=T

`q_2=m_2c_2* (T-T_3)`

`-q_1=q_2` (Law of Conservation of energy)

`-(m_1* c_1* (T-T_1))=m_2* c_2* (T-T_3)`

On substituting all values:

`-(18.5 g* 0.21 J/g^oC* (T-97.38^oC))=75.7 g* 4.184 J/g^oC* (T-21.52 ^oC)`

we get, T = 22.44°C

The final temperature of the water is 22.44°C.