问题 问答题

概念建构探究:不同物质传导热的能力不同,物质传导热的能力可用导热系数λ来表示.

导热系数是热的传导作用在1平方厘米截面上1秒钟内温差为1℃时通过长度1厘米的热量.

下面是一些物质的导热系数(J/cm•℃•s)

物质名称导热系数物质名称导热系数
  银  4.074  玻璃  0.0084
  铜  3.864  水  0.00596
  铝  2.100  冰  0.02310
  钨  2.016  空气  0.000239
  铁  0.672  棉花  0.000588
(1)由上述表格看出,金属的导热系数比非金属物质的导热系数______.

(2)导热系数越大的物质,传导热的能力越______.

(3)请写出导热的系数的数学表达式,并说明其中各个字母分别表示什么量:

①表达式:______;②各个字母所表示的量:______.

(4)计算题:将一块长为20cm、截面积为0.5cm2的铝片加热到100℃后,将一端接触一杯质量为500g、初温为20℃的水,热传递结束时水温为30℃,求经过多长时间铝片与水之间的热传递就结束了?(不计热量损失)

答案

(1)由表格中数据得出,金属的导热系数比非金属物质的导热系数大;

(2)导热系数越大,说明在1cm2截面上1s内温差为1℃时通过长度1cm的热量越大,即传导热的能力越强;

(3)①由导热系数是指在1cm2截面上1s内温差为1℃时通过长度1cm的热量可得关系式:λ=

QL
St△T

②上式中:λ为导热系数,Q为传递的热量;L为传导的长度,S为物体的横截面积,t为传热的时间,△T为物体的温差;

(4)不计热量损失,Q=Q水吸=cm(t-t0)=4.2×103J/(kg•℃)×0.5kg×(30℃-20℃)=2.1×104J,

∵λ=

QL
St△T

∴t=

QL
Sλ△T
=
2.1×104J×20cm
0.5cm2×2.1J/(cm•℃•s)×70℃
≈5714s.

故答案为:(1)大;(2)强;(3)λ=

QL
St△T
;(4)经过5714s铝片与水之间的热传递结束.

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