How much heat energy is gained when 5 kg of water at 20c is brought to its boiling point?
Answers
Given,
Mass of the substance( water ) = 5 kg
= 5 x 1000 g
= 5000 g
∴ Mass of water = 5000 g
Change in temperature = temperature at boiling pt. - 20°C
= 100°C - 20°C
= 80°C
∴ Change in temperature = 80°C
We know that the specific heat capacity of water is 1 Calorie / g °C
Now,
Heat given to the water,
Q = m c Δt
In the question,
m = 5000g , c = 1 calorie / g °C , Δt = 80°C
∴ Q = 5000 x 1 x 80 x g x calorie x 1°C / g °C
Q = 5000 x 1 x 80 calorie
Q = 5 x 80 k calorie
Q = 400 k calorie
Therefore, the 400 k calorie heat energy is gained when 5 kg of water at 20c is brought to its boiling point.
Given,
Mass of water (m) = 5kg = 5 × 10³ gram
Initial Temperature (t₁) = 20°C
We know that boiling point of water is 100°C, We have brought it to 100°C
∴ Final Temperature (t₂) = 100°C
Temperature Difference(Δt) = t₂ - t₁
= 100 - 20
= 80°C
Heat needed in this process (Q)= msΔt
Where, s = 1 calorie/gram, which is the specific heat of water. Taking m in gram and t in °C and s in cal./g, Unit of Q will be in calorie.
Q = msΔt
Q = 5 × 10³ × 1 × 80
Q = 4 × 10⁵ calorie.
Q = 400 kilo calorie