Determine the Specific Heat Capacity (Cs) of the liquid. Where: Mass of liquid (M₂) = 0.0012Kg, Initial temperature (₁)=23°C, Final temperature (9₂) = 62°C, Voltage (V)= 8v, Current (1) = 0.95A, Time (t) = 270s, Specific heat capacity of liquid (C₁) = ? 1
The Specific Heat Capacity of the liquid is 43.846 KJ/(Kg K).
It is given that Mass of liquid (M) = 0.0012Kg, Initial temperature (T1) = 23°C, Final temperature (T2) = 62°C, Voltage (V) = 8v, Current (I) = 0.95A, Time (t) = 270s
The quantity of heat that must be applied to an object in order to cause a unit change in temperature is known as the heat capacity or thermal capacity of that object.
We know that heat capacity for a substance is :
H = m*C*ΔT - equation (1)
When a conductor is subjected to current flow, the conductor's free electrons are set in motion and collide with one another. Moving electrons experience kinetic energy loss and partial thermal energy conversion as a result of the collision. This impact of current is referred to as its heating effect.
Due to electric current, heat energy is :
H = Power * Time
H = Current * Voltage * Time
H = I*V*t - equation (2)
Using equation (1) and (2),
m*C*ΔT = I*V*t
Substituting the values for m, ΔT, I, V and t.
0.0012Kg * C * (62 - 23)K = 0.95A * 8v * 270s
Solving for C, we get C = 43.846 KJ/(Kg K)
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