Question

A 300 g glass thermometer initially at 32 ◦C is put into 157 cm3 of hot water at 95 ◦C. Find the final temperature of the thermometer, assuming no heat flows to the surroundings. The specific heat of glass is 0.2 cal/g · ◦ C and of water 1 cal/g · ◦ C.

Answer 1

T_f= 77.58° C

Step-by-step explanation:

from simple calorimetry we can write that

`Q_w = m_wc_w(T_f-T_w)`

and

`Q_g = m_gc_g(T_f-T_w)`

Where

Q_w = heat content of water

Q_g= heat content of glass

m_g= mass of glass

m_w= mass of water

T_f= final temp

T_w= temp of water

T_g= temp of glass

m_w =mass of water

m_g= mass of glass

The specific heat of glass is 0.2 cal/g · ◦ C and of water 1 cal/g · ◦ C.

Now in given case

Q_w+Q_g=0

therefore

`Q_w = m_wc_w(T_f-T_w)`+
`Q_g = m_gc_g(T_f-T_w)`=0

⇒T_f=
`(m_gc_gT_g+m_wc_wT_g)/(m_wc_w+m_gc_g)`

putting values we get

T_f=
`(300*0.2*32+157*1*95)/(157*1+300*0.2)`

T_f= 77.58° C