Correction: The temperature change is from 20 °C to 30 °C.
Answer:
Cp = 1.0032 J.g⁻¹.°C⁻¹
Solution:
The equation used for this problem is as follow,
Q = m Cp ΔT ----- (1)
Where;
Q = Heat = 5016 J
m = mass = 500 g
Cp = Specific Heat Capacity = ??
ΔT = Change in Temperature = 30 °C - 20 °C = 10 °C
Solving eq. 1 for Cp,
Cp = Q / m ΔT
Putting values,
Cp = 5016 J / (500 g × 10 °C)
Cp = 1.0032 J.g⁻¹.°C⁻¹
Sir William Alexander Clarke Bustamante.
Hope This Helps!
It can be done. Normally the boiling point of water is 100°C. It will boil at temperature greater than 100°C more quickly. Water can be boiled at 95°C but for that the atmospheric pressure of the water should be decreased which will decrease the boiling point of water.
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Concept :</h3>
To boil water at 95°C, decrease the atmospheric pressure.
At 105°C, the water will be boiling quickly than normal at 100°C.
Answer:
1 liter (L) = 1000 milliliters (mL)
Explanation:
