Question

A 0.350 kg block at -27.5 degrees c is added to 0.217 kg of water at 25 degrees c. they come to equilibrium at 16.4 degrees c what is the specific heat of the block

Answer 1

Step-by-step explanation:

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)