Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
10,000J will raise the temperature of 1 liter of water by 2.38 °C
Using Q = mcΔT where Q = quantity of heat supplied = 10000 J,
m = mass of 1 liter of water = ρV where ρ = density of water = 1000 kg/m³ and V = volume of water = 1 liter = 1 dm³ = 1 × 10⁻³ m³.
So, m = ρV = 1000 kg/m³ × 1 × 10⁻³ m³ = 1 kg,
c = specific heat capacity of water = 4200 J/kg°C and ΔT = temperature change.
Making ΔT subject of the formula, we have
ΔT = Q/mc
Substituting the values of the variables into the equation, we have
ΔT = Q/mc
ΔT = 10000 J/(1 kg × 4200 J/kg°C)
ΔT = 10000 J/(4200 J/°C)
ΔT = 2.38 °C
So, 10,000J will raise the temperature of 1 liter of water by 2.38 °C
Learn more about temperature change here:
https://brainly.com/question/16384350
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.