The heat cause 300g water temperature increase from 20 to 26 celcius. The heat transferred would be: 300g * (26 °C -20 °C) *4.2 joule/gram °C= 7560J
The unknown substance is added to the water, so its final temperature should be the same as the water. The calculation would be:
7560J= 124g * (100-26)* specific heat
specific heat= 7560J / 124g / 74 °C= 0.8238 J/gram °C
I don’t know dawg but I’m just answering this so I can complete the steps good luck tho
I think is 1 and a half km
They are too small to see with the naked eye
Answer:
it will a i did the quiz got it all right
Explanation:
