Answer:
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Explanation:
<u>Step 1: </u>Data given
mass of water = 300 grams
initial temperature = 10°C
final temperature = 50°C
Temperature rise = 50 °C - 10 °C = 40 °C
Specific heat capacity of water = 4.184 J/g °C
<u>Step 2:</u> Calculate the heat
Q = m*c*ΔT
Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)
Q = 50208 Joule = 50.2 kJ
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
17, to be neutral it has to have equal number of protons and electrons
Answer:
The final temperature is 39.58 degree Celsius
Explanation:
As we know
Q = m * c * change in temperature
Specific heat of water (c) = 4.2 joules per gram per Celsius degree
Substituting the given values we get -
5750 = 335 * 4.2 * (X - 35.5)
X = 39.58 degree Celsius
