Question

how much energy is required to heat 0.24 kg lutetium from 296.2k to 373.5k? the specific heat for lutetium is 0.154 j/g·k. a 285.7 joules b 3694 joules c 115.8 joules d 2857 joules

Explanation:

Step1: Convert mass from kg to g

Given mass \(m = 0.24\space kg=0.24\times1000\space g = 240\space g\)

Step2: Calculate temperature change

\(\Delta T=T_{final}-T_{initial}=373.5 - 296.2=77.3\space K\)

Step3: Use heat formula \(Q = mc\Delta T\)

Here \(c = 0.154\space J/g\cdot K\), \(m = 240\space g\), \(\Delta T=77.3\space K\)
\(Q=(240)\times(0.154)\times(77.3)\)
\(Q = 240\times0.154\times77.3=240\times11.9042 = 2857.008\approx2857\space J\)

Answer:

D. 2857 joules