In an experiment for the specific heat of a metal a `0.20 kg` block of the metal at `150 .^(@)C` is dropped in a copper calorimeter (of water equivale

Question

In an experiment for the specific heat of a metal a `0.20 kg` block of the metal at `150 .^(@)C` is dropped in a copper calorimeter (of water equivale

In an experiment for the specific heat of a metal a `0.20 kg` block of the metal at `150 .^(@)C` is dropped in a copper calorimeter (of water equivalent `0.025 kg` ) containing `150 cm^(3)` of water of `27^(@)C`. The final temperature is `40 .^(@)C`. Calculate the specific heat of the metal:
A. `434.3 J//kg- .^(@)C`
B. `222.6 J//kg-.^(@)C`
C. `167.3 J//kg- .^(@)C`
D. `712.5 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

Correct Answer - A
Let specific heat of the metal is s, the heat lost by metal block `= ms Delta theta`
`= 0.02 s (150 - 40) = 22 s`
Mass of watr in the calorimeter `= rho V`
`= 1000 xx (150 xx 10^(-6)) = 0.15 kg`
Heat gained by water and calorimeter
`m_(w) s Deltha theta + W Delta theta`
`= (0.15 xx 4200 + 0.025 xx 4200) xx (40 - 27)`
`735 xx 13 = 9555 J`
By principle of calorimetry, we have
`22s = 9555`
`:. s = (9555)/(22) = 434.3 J//kg - .^(@)C`