Question

answer these question in detail​

Answer 1

Answer:

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 gain 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 is 45.5°C

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