Temperature 7595

Colorless liquid weighing m = 200 g is heated with constant stirring on a stove with power input P0 = 600W. 80% of the supplied energy is used to heat the liquid. Selected measured values of liquid temperature as a function of time are recorded in the table:

τ [s]	0	10	20	30	40	50	60
t [°C]	18,0	23,7	29,4	35,1	40,8	46,5	52,2

a) Draw a graph of the dependence of the liquid temperature on time and determine when it will be
have a liquid temperature of 30°C.
b) What will the temperature of the liquid be after 70 seconds?
c) In the same graph, record the dependence of temperature on time if we increase the amount of liquid 1.5 times
d) What is the specific heat capacity of the liquid?
During heating, the liquid does not evaporate or boil.

Correct answer:

t₃₀ = 26.5 s
b = 57.9 °C
t₅ = 34.8 °C
c = 4210.5263 J/kg/°C

Step-by-step explanation:

T (20 s) = 20.4 ° C T (30 s) = 35.1 ° C T (t) = a_{x} + b 20.4 = a \cdot 20 + b 35.1 = a \cdot 30 + b 20 a + b = 20.4 30 a + b = 35.1 P i v o t : R o w 1 \leftrightarrow R o w 2 30 a + b = 35.1 20 a + b = 20.4 R o w 2 - \frac{2 0}{3 0} \cdot R o w 1 \to R o w 2 30 a + b = 35.1 0.33 b = - 3 b = \frac{- 3}{0 . 3 3 3 3 3 3 3 3} = - 9 a = \frac{3 5 . 1 - b}{3 0} = \frac{3 5 . 1 + 9}{3 0} = 1.47 a = \frac{1 4 7}{1 0 0} = 1.47 b = - 9 t_{30} = (30 - b) / a = (30 - (- 9)) / 1.47 = \frac{1 3 0 0}{4 9} s = 26.5 s

m = 200 g \to kg = 200 : 1000 kg = 0.2 kg P_{0} = 600 W P_{1} = 0.80 \cdot P_{0} = 0.80 \cdot 600 = 480 W t = 60 - 50 = 10 s d = (52.2 - 46.5) / t = (52.2 - 46.5) / 10 = \frac{5 7}{1 0 0} = 0.57 ° C/s b = 52.2 + (70 - 60) \cdot d = 52.2 + (70 - 60) \cdot 0.57 = \frac{5 7 9}{1 0} ° C = 57.9 ° C

m_{1} = 1.5 \cdot m = 1.5 \cdot 0.2 = \frac{3}{1 0} = 0.3 kg t_{0} = 18 ° C t_{1} = 23.7 / 1.5 = \frac{7 9}{5} = 15.8 ° C t_{2} = 29.4 / 1.5 = \frac{9 8}{5} = 19.6 ° C t_{3} = 35.1 / 1.5 = \frac{1 1 7}{5} = 23.4 ° C t_{4} = 40.8 / 1.5 = \frac{1 3 6}{5} = 27.2 ° C t_{5} = 46.5 / 1.5 = \frac{1 7 4}{5} = 34.8 ° C t_{5} = 52.2 / 1.5 = \frac{1 7 4}{5} ° C = 34.8 ° C

Q = P_{1} \cdot t = 480 \cdot 10 = 4800 J Q = m c (t_{1} - t_{0}) c = \frac{Q}{m \cdot ( 2 3 . 7 - 1 8 )} = \frac{4 8 0 0}{0 . 2 \cdot ( 2 3 . 7 - 1 8 )} = \frac{8 0 0 0 0}{1 9} J/kg/ ° C = 4210.5263 J/kg/ ° C

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