An electrical heater is used to supply 160.0 J of energy to
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An electrical heater is used to supply 160.0 J of energy to a 30.0 g sample of water,originally at 300.0 K. Determine the final temperature in degrees Kelvin.(the specific heat capacity for water is 75.291 Jmol-1K-1)
An electrical heater is used to add 19.50 kJ of heat to a constant-volume calorimeter. The temperature of the calorimeter increases by 4.48°C. When 2.25 g of propanol (C3H7OH) is burned in the same calorimeter,the temperature increases by 17.38°C. Calculate the molar heat of combustion for propanol (enter in kJ).
A home swimming pool contains 150 m3 of water. At the beginning of the swimming season,the water must be heated from 15 to 28°C.How many joules of heat energy must be supplied?
Suppose 100.0 mL of 1.50 M HCl and 100.0 mL of 1.50 M NaOH,both initially at 21.0°C,are mixed in a thermos flask. When the reaction is complete,the temperature is 31.2°C. Assuming that the solutions have the same heat capacity as pure water,compute the heat released (in kJ).
A coin dealer,offered a rare silver coin,suspected that it might be a counterfeit nickel copy. The dealer heated the coin,which weighed 17.5 g to 100°C in boiling water and then dropped the hot coin into 20.5 g of water at T = 16.5°C in an insulated coffee-cup,and measured the rise in temperature. If the coin was really made of silver, what would the final temperature of the water be (in °C)?(for nickel,s = 0.445 J/g°C; for silver,s = 0.233 J/g°C )
能做出几道做几道,
An electrical heater is used to add 19.50 kJ of heat to a constant-volume calorimeter. The temperature of the calorimeter increases by 4.48°C. When 2.25 g of propanol (C3H7OH) is burned in the same calorimeter,the temperature increases by 17.38°C. Calculate the molar heat of combustion for propanol (enter in kJ).
A home swimming pool contains 150 m3 of water. At the beginning of the swimming season,the water must be heated from 15 to 28°C.How many joules of heat energy must be supplied?
Suppose 100.0 mL of 1.50 M HCl and 100.0 mL of 1.50 M NaOH,both initially at 21.0°C,are mixed in a thermos flask. When the reaction is complete,the temperature is 31.2°C. Assuming that the solutions have the same heat capacity as pure water,compute the heat released (in kJ).
A coin dealer,offered a rare silver coin,suspected that it might be a counterfeit nickel copy. The dealer heated the coin,which weighed 17.5 g to 100°C in boiling water and then dropped the hot coin into 20.5 g of water at T = 16.5°C in an insulated coffee-cup,and measured the rise in temperature. If the coin was really made of silver, what would the final temperature of the water be (in °C)?(for nickel,s = 0.445 J/g°C; for silver,s = 0.233 J/g°C )
能做出几道做几道,
第一题:30g水即1.67摩尔,利用物理学公式:温度变化=能量×摩尔质量/比热容即可计算得出变化的热力学温度值,再将热力学温度转换到Kelvin温度.
第二题:升温4.48摄氏度需要19.5kJ,可以用等比公式计算出升温17.38摄氏度所需热量,用所得热量值除以2.25g酒精的摩尔质量即可得出答案.
第三题:纯粹物理题,热量=质量×比热容(第一题有)×温度变化
第四题:同第3题,主要问题在于求质量,100mL×1.5mol/L科德其摩尔质量
第五题:
设最终平衡度为x,(x-16.5)×水比热×20.5=(100-x)×银比热×17.5,得出如果是银币的最终温度,再用同一种方法求镍的最终温度,进行比较
再问: 第一二题懂了。谢谢。 其他的算完都显示不正确 第三题我算是1.47e5 第四题我算是6.406 第五题我算是96.1。我不知道哪里有算错。你能帮我算吗
再答: 第一题的水比热质量单位不是按克而是按摩尔计,你要好好计算清楚,水18g=1mol
再问: 第三题还是算不对……
再答: 你说说你的思路,估计里面有点小问题
再问: 150立方米=150000升,用密度算求出水的质量是150000000克,热量=150000000*75.291*13=1.47e11 可是显示结果不对
再答: 都说你那个质量没有换成摩尔量咯呵呵~~
再问: 啊~会了会了。谢谢! 还有,第四题,先算摩尔n=0.1*1.5=0.15,热量=0.15*10.2*75.291=115.2J=0.1152kJ,错误在哪里?结果显示不正确呃~
再答: 200ml啊,被你喝了一半啦!
再问: 0.2*0.15*10.2*75.291=0.23KJ可是还是不对呢,奇怪呃…
再答: 不用乘以0.15的啊
再问: 不对哦,这个答案还是不正确呃…
再答: 那个比热容是相对摩尔的,你把0.2L换算成摩尔了没!
再问: 想要换成摩尔不是应该体积乘以浓度嘛?就是0.2*1.5呀…
再答: 人家题目不用你管盐酸和碱的摩尔量,就是200ml当作是纯水,从21度上升到31.2度,要你计算有多少热量产生了。 所以你要做的是把200ml纯水有多少摩尔量算出来~~ 做完后你得采纳啊,一道题搞了这么多文字……
第二题:升温4.48摄氏度需要19.5kJ,可以用等比公式计算出升温17.38摄氏度所需热量,用所得热量值除以2.25g酒精的摩尔质量即可得出答案.
第三题:纯粹物理题,热量=质量×比热容(第一题有)×温度变化
第四题:同第3题,主要问题在于求质量,100mL×1.5mol/L科德其摩尔质量
第五题:
设最终平衡度为x,(x-16.5)×水比热×20.5=(100-x)×银比热×17.5,得出如果是银币的最终温度,再用同一种方法求镍的最终温度,进行比较
再问: 第一二题懂了。谢谢。 其他的算完都显示不正确 第三题我算是1.47e5 第四题我算是6.406 第五题我算是96.1。我不知道哪里有算错。你能帮我算吗
再答: 第一题的水比热质量单位不是按克而是按摩尔计,你要好好计算清楚,水18g=1mol
再问: 第三题还是算不对……
再答: 你说说你的思路,估计里面有点小问题
再问: 150立方米=150000升,用密度算求出水的质量是150000000克,热量=150000000*75.291*13=1.47e11 可是显示结果不对
再答: 都说你那个质量没有换成摩尔量咯呵呵~~
再问: 啊~会了会了。谢谢! 还有,第四题,先算摩尔n=0.1*1.5=0.15,热量=0.15*10.2*75.291=115.2J=0.1152kJ,错误在哪里?结果显示不正确呃~
再答: 200ml啊,被你喝了一半啦!
再问: 0.2*0.15*10.2*75.291=0.23KJ可是还是不对呢,奇怪呃…
再答: 不用乘以0.15的啊
再问: 不对哦,这个答案还是不正确呃…
再答: 那个比热容是相对摩尔的,你把0.2L换算成摩尔了没!
再问: 想要换成摩尔不是应该体积乘以浓度嘛?就是0.2*1.5呀…
再答: 人家题目不用你管盐酸和碱的摩尔量,就是200ml当作是纯水,从21度上升到31.2度,要你计算有多少热量产生了。 所以你要做的是把200ml纯水有多少摩尔量算出来~~ 做完后你得采纳啊,一道题搞了这么多文字……
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