Answer:20,930 J
Step-by-step explanation:
20,930 J
When 200 grams of water cools from 50°C to 25°C, the total amount of heat energy released is 20,930 J12. The formula to calculate heat energy is q = mCΔT, where q is the heat energy, m is the mass of the substance, C is the specific heat, and ΔT is the change in temperature1. The specific heat of water is 4.186 J/g °C1.
21 kJ
Step-by-step explanation:
The formula for the heat q is
q = mCΔT
Data:
m = 200 g; C = 4.184 J·°C⁻¹g⁻¹
T₁ = 50. °C; T₂ = 25 °C
Calculations:
ΔT = T₂ - T₁
ΔT = 25 – 50.
ΔT = -25 °C
q = 200 × 4.184 × (-25)
q = -21 000 J Convert to kilojoules
q = -21 kJ
The negative sign shows that energy is released, so the water has released 21 kJ of energy.
Answer : The heat released by the eater is, 2.1times 10^4J[/tex]
Explanation :
Formula used :
Q=mtimes ctimes Delta T
or,
Q=mtimes ctimes (T_2-T_1)
where,
Q = heat released = ?
m = mass of water = 200 g
c = specific heat of water = 4.184J/g^oC
T_1 = initial temperature = 50^oC
T_2 = final temperature = 25^oC
Now put all the given value in the above formula, we get:
Q=200gtimes 4.184J/g^oCtimes (25-50)^oC
Q=20920J=2.1times 10^4J[/tex]
Therefore, the heat released by the eater is, 2.1times 10^4J[/tex]
Answer : The heat released by the eater is, 2.1times 10^4J[/tex]
Explanation :
Formula used :
Q=mtimes ctimes Delta T
or,
Q=mtimes ctimes (T_2-T_1)
where,
Q = heat released = ?
m = mass of water = 200 g
c = specific heat of water = 4.184J/g^oC
T_1 = initial temperature = 50^oC
T_2 = final temperature = 25^oC
Now put all the given value in the above formula, we get:
Q=200gtimes 4.184J/g^oCtimes (25-50)^oC
Q=20920J=2.1times 10^4J[/tex]
Therefore, the heat released by the eater is, 2.1times 10^4J[/tex]
Specific Heat:
Heat Energy= Mass of substance X Specific Heat X Change in Temp.
1. change in temp |50-25| = 25
2. specific heat of Water(H2O) = cal/g (Celsius) 1.000
heat energy= 200g X 1.000 X 25
Heat energy = 5000cal