Question

A thermometer of mass 0.055 kg and heat capacity 46.1 J/K reads 15.0°C. It is then completely immersed in 0.300 kg of water and it comes to the same final temperature as the water. If the thermometer reads 44.4°C, what was the temperature of the water berfore insertion of the thermometer, neglecting other heat losses?

Answer 1

mass of thermometer = 0.055 kg
heat capacity = 46.1 J/K
initial temperature = 15° C
final temperature = 44.4 ° C

mass of water = 0.300 kg
heat capacity of water = 4.2 J/K
initial temperature = T ( let )
final temperature =44.4° C

now,
use concept of calorimetry ,
Heat loss = Heat gain

here water is at higher temperature , so heat loss by water . and heat again by thermometer.

heat loss by water = Mw × Cw × ( Ti - 44.4)
= 0.3 × 4190 × ( Ti - 44.4)
[ Cw = 4190 j/kg.K ]

heat gain by thermometer = Mt× (St/Mt) × ( 44.4° - 15°) = St × 29.4 °

[ ( St/Mt ) = 46.1 J/K is known as heat capacity ]

hence,
0.3 × 4190 × ( Ti -44.4°) = 46.1 × 29.1

( Ti -44.4°) = 46.1 × 29.4/0.3 × 4190

( Ti - 44.4°) = 1.07° C
Ti = 44.4 + 1.07 = 45.47°C ≈ 45.5° C

hence, water insertion temperature = 45.5°C

Answer 2

The heat transfers for the water Q w is, Q w = m w c w ( T f – T i) Here, mass of water is m w, specific heat capacity of water is c w, final temperature is T f and initial temperature is T i. The heat transfers for the thermometer Q t is Q t = C t? T t Here, heat capacity of thermometer is C t and ? T t is the temperature difference. As the internal energy of the system is zero and there is no work is done, therefore substitute ? E int = 0 and W = 0 in the equation Q + W = ? E int, Q + W = ? E int Q + 0= 0 So, Q = 0 Or, Q w + Q t = 0 m w c w ( T f – T i)+ C t? T t = 0 So, T i = ( m w c w T f + C t? T t )/ m w c w Here ? T t = 44.4 ° C - 15.0 ° C = 29.4 ° C To obtain the temperature of the water...