PHYS 223 - Lab Report 6 (Mechanical Equivalence of Heat)

PHYS 223 Mechanical Equivalence of Heat Megan Cousins Tuesday-Thursday Objective Our goal is to find the conversion rate between Joules and Calories through the electrical method of mechanical equivalence of heat specifically in a container of water. Introduction For this experiment we find most of our experimental values using calories, defined as the amount of heat energy necessary to raise the temperature of one gram of pure water one degree Celsius. In the experiment, an electric coil is used to raise the temperature of a known mass of water a measured number of degrees. The coil is immersed in the water and the current and voltage of the coil is set and recorded. Questions 1. Calculate W, Q t and K for both runs. W = 11.74 Watts Q t = 2739.35 calories K = 6.77 J 2. Discuss the principal sources of error in this experiment and justify the difference between your value of the mechanical equivalent of heat and the correct value. The principal sources of error in this experiment are not properly measuring the time that it takes in order to heat the water to the difference that we need, as well as the property that the water continues to warm in temperature after the electrical current has been turned off. Another source of error would be the transfer of heat into other elements such as the surrounding air. The difference between our calculated value and the correct value could be justified in the idea that we initially forgot to start the timer as soon as the electrical current was turned on. 3. Using your data, calculate how long it would take to heat the water from 20 °C to 100 °C by sending a current of 2 amperes through the same coil immersed in the water.

K = W/Q t K=W/(M c C c ∆T) +(M w C w ∆T) K = (V I) t / (M c C c ∆T) + (M w C w ∆T) 6.77 = (5.7 x 2) (t) / (29.4 x 0.22 x 14.1) + (190.8 x 14.1) 6.77 = (11.4) (t) / (91.20) + (2288.25) 6.77 = (11.4) (t) / 2739.35 18,545.40 = 11.4 (t) 1,626.79 seconds = t 4. In question 3, how long would it take if a current of 4 amperes were used? K = W/Q t K=W/(M c C c ∆T) +(M w C w ∆T) K = (V I) t / (M c C c ∆T) + (M w C w ∆T) 6.77 = (5.7 x 4) (t) / (29.4 x 0.22 x 14.1) + (190.8 x 14.1) 6.77 = (22.8) (t) / (91.20) + (2288.25) 6.77 = (22.8) (t) / 2739.35 18,545.40 = 22.8 (t) 813.39 seconds = t 5. What must be the resistance of a coil of wire in order that a current of 2 amperes flowing through the coil may produce 1000 calories of heat per minute? Ω = P / I 2 P = V I W = P t W = K Q t (V I) t = K Q (V (2)) 60 = 4.184 (1000) (V (2)) 60 = 4,184 V = 34.87 Volts P = 69.73 Watts Ω = 69.73 / 2 2