Question

A 15.75g PIECE OF IRON ABSORBS 1086.75 JOULES OF HEAT ENRGY,AND ITS TEMPERATURE CHANGES FROM 25C TO 75C. CALCULATE THE SPECIFICHEAT CAPACITY OF IRON.

Answer 1

A 15.75g PIECE OF IRON ABSORBS 1086.75 JOULES OF HEAT ENERGY, AND ITS TEMPERATURE CHANGES FROM 25C TO 75C. CALCULATE THE SPECIFIC HEAT CAPACITY OF IRON.

Solution.

$$Q = c * m * (t _ {2} - t _ {1}) \quad \text{-HEAT ENERGY ABSORBED BY PIECE OF IRON}$$ $$c = \frac {Q}{m * (t _ {2} - t _ {1})} = \frac {1 0 8 6 . 7 5 \text{ joules}}{0 . 0 1 5 7 5 k g * (7 5 {}^ {\circ} \mathrm {C} - 2 5 {}^ {\circ} \mathrm {C})} = 1 3 8 0 \frac {\text{joules}}{k g * {}^ {\circ} \mathrm {C}} \quad \text{-THE SPECIFIC HEAT CAPACITY OF IRON}$$