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

Answer:

T_f= 77.58° C

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= $\frac{m_gc_gT_g+m_wc_wT_g}{m_wc_w+m_gc_g}$

putting values we get

T_f= $\frac{300\times0.2\times32+157\times1\times95}{157\times1+300\times0.2}$

T_f= 77.58° C