When 90 kJ is removed from a 2-kg copper bar, its temperature drops from 200^$\circ$C to 85^$\circ$C.The specific heat of iron is A)0.16 kJ/kg.^$\circ$C. B)0.19 kJ/kg.^$\circ$C. C)0.23 kJ/kg.^$\circ$C. D)0.39 kJ/kg.^$\circ$C.

Question

Quizplus · Accepted Answer

The specific heat capacity can be calculated using the formula: $ Q = mc\Delta T $. Rearranging for $ c $, we have $ c = \frac{Q}{m\Delta T} $. Substituting the given values: $ c = \frac{90 \text{ kJ}}{2 \text{ kg} \times (200 - 85) \text{ °C}} = 0.19 \text{ kJ/kg.°C} $.

When 90 KJ Is Removed from a 2-Kg Copper Bar ∘\circ∘

When 90 KJ Is Removed from a 2-Kg Copper Bar $\circ$