Question

If a heating coil of the R=2.5Ω is switched on with a I = 5A for 8 minutes to heat up a 2 kg liquid, its temperature rises from T= 20^oC to 55^oC. If we assume that there is no phase change, no heat loss and the heat absorbed by the heating coil is 280J, what is the liquid specific heat capacity C (J/kg/K))?

Select one:

a. 428.6

b. 3.14

c. 270.2

d. 424.6

Answer 1

Power generated by coil = i² * Resistance * time(s)

power = 5^2 * 2.5 * 8 * 60

power = 30000 Joule

now coil absorb 280 J of energy so energy left for liquid = 30000 - 280

E = 29720 Joule

now this enegy will be equal to melting energy required for liquid

so E = m * C * delta_T

29720 = 2 * C * 35

C = 424.57   

C = 424.6 (j/kg-k) <<<<<<<<<OPTION D