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A 300 g glass thermometer initially at 23 ◦C is put into 236 cm3 of hot water at 87 ◦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.

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Answer:


74^(\circ) C

Step-by-step explanation:

We are given that

Mass of glass,
m=300 g


T_1=23^(\circ)

Volume,
V=236cm^3

Mass of water=
density* volume=1* 236=236 g

Density of water=
1g/cm^3

Temperature of hot water,
T=87^(\circ)

Specific heat of glass,
C_g=0.2cal/g^(\circ)C

Specific heat of water,
C_w=1 cal/g^(\circ)C


Q_(glass)=m_gC_g(T_f-T_1)=300* 0.2(T_f-23)


Q_(water)=m_wC_w(T_f-T)=236* 1(T_f-87)


Q_(glass)+Q_(water)=0


300* 0.2(T_f-23)+236* 1(T_f-87)


60T_f-1380+236T_f-20532=0


296T_f=20532+1380=21912


T_f=(21912)/(296)=74^(\circ) C

User Sangram Anand
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