采用传统高温固相反应法制备La1.2-xTbxSr18Mn2O7(x=0,0.05)多晶样品,利用振动样品磁强计(VSM)测量了样品在不同温度下的磁化强度随外场变化曲线(M-H),然后利用正交多项式最小二乘拟合方法对钙钛矿锰氧化物La1.2-xTbxSr18Mn2O7(x=0,0.05)的磁化强度曲线进行拟合,再根据磁熵变的热力学公式计算出样品的磁熵变值.通过计算得到La1.2-xTbxSr18Mn2O7(x=0,0.05)多晶样品的居里温度分别为123 K和75 K,在外加磁场20 kOe下x=0样品和x=0.05样品的最大磁熵变值分别为2.26 J/(kg·K)和1.6 J/(kg·K),所以x=0样品可作为高温区(77 K以上)磁制冷材料,而x=0.05样品可作为中温区(20 K～77 K)磁制冷材料.计算结果显示,拟合数据和实验数据非常接近,结果比较满意,说明该方法适用于La1.2-xTbxSr18Mn2O7(x=0,0.05)多晶样品的磁熵变计算.
The polycrystalline samples of double-layered magnetite La1.2-xTbxSr1.sMn2O7 (x =0,0.05) were prepared by the traditional solid state reaction method.The magnetization curves were measured by a vibrating sample magnetometer (VSM),from which the function M(H,T) was obtained using orthogonal polynomial least square method fitting.Magnetic entropy changes can be calculated according to the magnetization curves.The results show that the Curie temperature and the maximum value of the magnetic entropy change under a 20 kOe magnetic field of La1.2-xTbxSr1.8Mn2O7 (x =0) is 123K and 2.26 J/(kg · K) respectively.The results show that the x =0 sample could be used as magnetic refrigerant material in the temperature region higher than 77 K.And the Curie temperature and the maximum value of the magnetic entropy change under a 20 kOe magnetic field of La1.2-xThxSr1.8Mn2O7 (x =0.05) is 75 K and 1.6 J/(kg · K) respectively.The results show that the x =0.05 sample could be used as magnetic refrigerant material in the temperature region of 20 K to 77 K.The results fit well with experimental data.This method could be used to calculate the magnetic entropy changes of our samples.
参考文献
- 下载量()
- 访问量()
- 您的评分：
-
10%
-
20%
-
30%
-
40%
-
50%