A 31.87 g sample of a substance is initially at 28.5 °C. After absorbing 1697 J...

A 31.87 g sample of a substance is initially at 28.5 °C. After absorbing 1697 J of heat, the temperature of the substance is 140.5 °C. What is the specific heat (c) of the substance?

Solutions

Expert Solution

Given, mass of sample (m)= 31.87 gm

Initial temperature (T₁) = 28.5⁰C

Final temperature (T₂) = 140.5⁰C

Hence, T = T₂ - T₁ = 140.5⁰C - 28.5⁰C = 112⁰C

Heat absorbed (q) = 1697 J

Specific heat (C) = ???

As we know,

q = m*C*T

C = q / m*T

C = 1697 J / 31.87 gm * 112⁰C

C = 1697 J / 3569.44 gm⁰C

C = 0.475 J/gm⁰C

queen_honey_blossom answered 1 year ago

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