Question

C.

A 35.5 g cube of aluminum initially at 48.5° C is submerged into 105.3 g of water at 15.4
What is the final temperature of both substances at thermal equilibrium?

Answer 1

The final temperature of both substances at thermal equilibrium is 17.3°C

Step-by-step explanation:

To calculate the final temperature of both substances at thermal equilibrium -:

First , we calculate the heat of A1 cube as follows -

q= mSΔT

(where q = heat of the cube , m = mass of cube , S= specific heat of cube {0.902j/g°C}, T = Temperature )

Putting the values given in the question ,

`q= 32.5g*0.902\j/g`°
`C`
`*(T_f-45.8`°
`C)`

`29.315*(T_f-45.8)J`

Now , calculate the heat of water -

q=mSΔT

Putting values from the question ,

`q=105.3g*4.18j/g`°
`C`
`* (T_f-15.4`°
`C)`

=
`440.154*(T_f-15.4)J`

Now ,

Heat lost by water A1= Heat gained by water [negative sign about heat lost]

`-29.315*(T_f-45.8)J`
`=440.154*(T_f-15.4)J`

`(-(T_f-45.8)J)/(T_f-15.4)J) =(440.154)/(29.315)=15.0`

`-T_f+45.8`°
`C=15T_f-231.2`°
`C`

`(45.8+231.2)`°
`C`=
`16T_f`

`16T_f=277.03`°
`C`

`T_f=(277.03)/(16)` = 17.3°C

Therefore , the final temperature of both substances at thermal equilibrium is 17.3°C