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In: Chemistry

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 2 years ago
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