Specific heat capacity laboratory - Assignment Example

substance by 1 degree which is obtained through the equation Q = m*C*∆T where ‘m’ refers to the mass of material, C is the specific heat and ∆T is the change of temperature. ∆T in Kelvins is the same in magnitude as ∆T in Celsius, hence, the specific heat for water may also be reported as 4.18 J/g-°C. Because the specific heat values for a given substance can vary slightly with temperature, the temperature is often precisely specified. (3) Record the initial temperature of water prior to heating.

Switch on the electric kettle and note down temperature reading measured after every 15 seconds until the instant is reached when the water starts boiling. The units for the special heat capacity based on C = Q / (m∆T) is J kg-1K-1 and by mere inspection, the rated power consumption of the kettle is 2200 Watts (or J·s-1). Thus, the maximum amount of energy that the kettle can be expected to transfer to the water over a period of 15 seconds is 33,000 Joules which is the product of power consumed and time elapsed.

Actual energy transfer may be less than this amount due to the energy lost through heating the kettle itself, the surrounding air, and even the noise which take up some dissipated heat in the process. Assuming complete energy transfer with a 100% efficient kettle, 2200*(t) = (200 grams)*(4.2 J g-1°C-1)*(100°C - 10°C) gives t = 34.36 seconds or the time it would take to get a 200-gram water to boil. The indicated form y = 0.462x + 27.34 translates to T = 0.462*t + 27.34. Clearly, the graph is a straight line, which plateaus at ~ 100 °C while the gradient remains constant (linear) with time until it reaches the boiling point.

Slope approaches zero or gradient flattens when the kettle is boiling. The gradient of this graph is the specific heat capacity of water 4610 J*kg-1K-1 or 4.61 J*g-1K-1(using the data points. Apparently, this makes an overestimate when compared to literature value of 4.2 J*g-1K-1 due to energy dissipation in the