Trần Phú Khang
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Một bếp điện khi hoạt động bình thường có điện trở R=80 ôm và cường độ dòng điện là 2,5A. A) tính nhiệt lượng mà bếp tỏa ra trong 1s. B)dùng bếp để đun sôi 1,5kg chất lỏng có nhiệt độ ban đầu là 20°c và nhiệt độ khi sôi là 100°c, thì thời hian đun sôi chất lỏng là 20 phút. Biết hiệu suất của bếp đạt 80%. Tính nhiệt lượng cần đun sôi lượng chất lỏng trên? C) tính nhiệt dung riêng của chất lỏng đó?
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1 câu trả lời 231
1 năm trước
a,
Nhiệt lượng mà bếp tỏa ra trong 1s là:
.80.1=500 J
b,
Nhiệt lượng mà bếp điện tỏa ra trong 20 phút là:
Qtp = 500 . 20 .60 = 600000 J
Hiệu suất của bếp là: 80%
Nhiệt lượng cần đun sôi lượng chất lỏng trên là:
600 000 x 80%= 480 000 (J)
c)
nhiệt dung riêng của chất lỏng đó là:
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