In this experiment voltage across the heater is `100.0V` and current is `10.0A` and heater was switched on for `t = 700.0sec` Initially elements were

Question

In this experiment voltage across the heater is `100.0V` and current is `10.0A` and heater was switched on for `t = 700.0sec` Initially elements were

In this experiment voltage across the heater is `100.0V` and current is `10.0A` and heater was switched on for `t = 700.0sec` Initially elements were at room temperature `theta_(0) = 10.0^(@)C` and final temperature was measured at `theta = 73.0^(@)C` Mass of empty calorimeter was `1.0kg` and the combined mass of calorimeter + liquid is `3.0kg` THe specirfice of the calorimeter `S_(c) = 3.0 xx 10^(3) J//kg^(@)C` The fall in temperature 350 second after switching off the heater was `7.0^(@)C` Find the specific heat capacity of the unknown liquid in proper significant
(A) `3.5 xx 10^(3) J//kg^(@)C` (B) `3.50 xx 10^(3) J//kg^(@)C` (C) `4.0 xx 10^(3) J//kg^(@)C` (D) `3.500 xx 10^(3) J//kg^(@)C` .

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

corrected final temperature `= theta_(f) = 73.0^(@) + 7.0^(@) = 80.0^(@)`
`S=(((100.0)(10.0)(700.0))/(80.0-10.00)(1.0)(3.0xx10^(3)))/(3.0-1.0)`
`=3.5 xx 10^(3)J//kg^(@)C` (According to addition and muktiplication rule of `S.F` ) .

4 views Answered 2023-02-05 15:29