Copper block of 3.1 kg is heated in a furnace from 20 degree Celsius to 520 degree Celsius and then placed on thick ice box. calculate specific heat of copper is mass of ice melted is 1.8 kg. (given= latent heat of fusion of ice = 333 kJ/kg.)
Answers
The specific heat capacity of copper is 0.386 kJ kg⁻¹ K⁻¹
• Given,
Mass of copper that is heated (m) = 3.1 kg
Initial temperature of the furnace (T1) = 20 °C
Final temperature of thr furnace (T2) = 520 °C
Therefore, change in temperature of the furnace (∆T) = T2 - T1 = 520 °C - 20 °C = 500 °C
• Now, as we know that a difference of 1°C is equal to 1 K, therefore, 500 °C = 500 K
Specific heat of the copper be (C) = ?
• Maximum heat that can be absorbed by copper (Q) = mC∆T
=> Q = 3.1 kg × C × 500 K -(i)
• Given, mass of ice melted by copper (m') = 1.8 kg
Latent heat of fusion of ice (L) = 333 kJ / kg
• Now, heat absorbed by ice = heat lost by copper = Q
Therefore, Q = m'L
Or, Q = 1.8 kg × (333 kJ / kg)
Or, Q = 599.4 kJ
• Putting the value of Q in equation (i), we get,
599.4 kJ = 3.1 kg × C × 500 K
=> C = 599.4 kJ / (3.1 kg × 500 K)
=> C = 0.386 kJ kg⁻¹ K⁻¹
• The specific heat capacity of copper is 0.386 kJ kg⁻¹ K⁻¹ .
Step-by-step explanation:
Hope it helps....Simple and easy answer
