A technician measures the specific heat of an unidentified liquid by immersing an electrical resistor in it. Electrical energy is converted to heat transferred to the liquid for 120 s at a constant rate of 65.0 W. The mass of the liquid is 0.780 kg, and its temperature increases from 18.55o C to 22.54o C. (a) Find the average specific heat of the liquid in this temperature range. Assume that negligible heat is transferred to the container that holds the liquid and that no heat is lost to the surroundings. (b) Suppose that in this experiment heat transfer from the liquid to the container or surroundings cannot be ignored. Is the result calculated in part (a) an ?overestimate or an underestimate of the average specific heat? Explain.

Solution 36E Step 1 of 7: (a) Find the average specific heat of the liquid in this temperature range. Assume that negligible heat is transferred to the container that holds the liquid and that no heat is lost to the surroundings. Given data, Power transferred, P= 65 W Time taken for transfer, t= 120 s Mass of the liquid, m= 0.780 kg Initial temperature of liquid, T = 18.55 c 0 i Final temperature of liquid, T = 22.54 c 0 f To find, Change in temperature, T = Energy transferred to liquid, E= Average specific heat of the liquid , c = Step 2 of 7: As we provide heat energy for the liquid, the liquid rises its temperature. Whereas in the given problem , the provided electrical energy from resistor is converted into heat energy, that is used by the liquid. The amount of energy dissipated by the resistor will be equal to the amount of energy used to increase the temperature of the liquid. The equation that relates change in temperature and heat gain or loss, Heat gain or loss is given by, Q= m c T ………………..1 Where Q is the heat gain or loss, m is the mass , c is specific heat andT is change in temperature. Step 3 of 7: To calculate energy dissipated by the resistor which is equal to heat energy gained by liquid, Using Power = Energy = E Time of transfert Solving for energy, E = P t Substituting P= 65 W or J/s and t =120 s E = (65 J/s)(120 s) E =7800 J Step 4 of 7: The same amount of energy will be utilized by the liquid to increase the temperature from T = i8.55 c to T = 22.f4 c, 0 Therefore change in temperature, T =22.54 c-18.55 c 0 T = 3.99º c