a 0.350 kg block at -27.5 (degree C) is added to 0.217 kg of water at 25.0 (degree Celsius). they come to equilibrium at 16.4 (degree Celsius). what is the specific heat of the block?

Answer 1

Heat = m*S* Change in temperature

Heat gained by block = Heat lost by water ( M = mass, S = Sp heat)

Mb *Sb * Change in temperature of block = Mw*Sw* change in temperature of water

0.35 * Sb *43.9 = 0.217* 4200*18.6

The change in temp of block is from -27.5 to zero to 16.4 = 43.9 ⁰C

Sw = 4200 J/degree C* Kg

So Sb = ( 0.217*4200*18.6) / ( 0.35 *43.9) J/⁰C*Kg = 1103.3 J/⁰C*Kg

Which is much lower than sp heat of water (Sw)