Answer:
25.2°C
Explanation:
Given parameters:
Energy applied to the water = 1000J
Mass of water = 50g
Final temperature = 30°C
Unknown:
Initial temperature = ?
Solution:
To solve this problem, we use the expression below:
H = m c Ф
H is the energy absorbed
m is the mass
c is the specific heat capacity
Ф is the change in temperature
1000 = 50 x 4.184 x (30 - initial temperature )
1000 = 209.2(30 - initial temperature)
4.78 = 30 - initial temperature
4.78 - 30 = - initial temperature
Initial temperature = 25.2°C
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Answer:
\large \boxed{\textbf{609 kJ}}
Explanation:
The formula for the heat absorbed is
q = mCΔT
Data:
m = 2.07 kg
T₁ = 23 °C
T₂ = 191 °C
C = 1.75 J·°C⁻¹g⁻¹
Calculations:
1. Convert kilograms to grams
2.07 kg = 2070 g
2. Calculate ΔT
ΔT = T₂ - T₁ = 191 - 23 = 168 °C
3. Calculate q
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