What Is the Specific Heat Capacity of the Liquid in This Thermodynamics Problem?

[dobedobedo]

To increase the temperature of a vessel with 1 K the amount of 50 J energy must be supplied. In the vessel 250 g of a liquid is added. With a heating spiral, which develops 15 W, the temperature is increased. Eventually it is stabilized and becomes constant until the heat supply gets turned off. Then the temperature decreases in the beginning with 1.2 K per minute. Calculate the specific heat capacity of the liquid.

How do I solve this rather basic problem? According to the textbook the answer should be 2.8 kJ/(kg*K)

Should I take into consideration that the vessel itself is heated, even though it's specific heat capacity and mass are unknown? I don't understand what facts I should ignore, and what facts I should take into consideration. All I know is that E = cmΔT and that heat is transferred from hot to cold. My attempt to a solution is that the supplied energy per second should be equal to the energy lost when the liquid is cooled. Out of this assumption I calculated a heat capacity of 3.0 kJ/(kg*K), which is rather close but still not correct.

Please help this lost soul.
)':

 

[ehild]

The heat capacity of the vessel C_vessel=50 J/K had to be taken into account. The heat added/removed is

Q=(c_fluid m_fluid+C_vessel)ΔT

ehild

 

[CWatters]

To work out the heat capacity of an object you need to know how much energy it takes to change the temperature by 1 degree.

You know how much energy it takes to raise the temperature of the vessel alone by 1k so you can work out it's heat capacity.

Then for the second step.. When the heating coil is first turned off you know the vessel is at an (unknown) temperature at which it looses 15W (15J/S). You know that because that's how much much power was required to maintain it at that temperature. You also know that when loosing 15J/S the temperature falls 1.2k/min. So you can work out the amount of energy needed to change the temperature of the combined vessel and liquid by 1K.

Then easy to calculate the heat capacity of the liquid alone by subtraction.

Then easy to convert that to the specific heat capacity (scale to 1kg).

 

[dobedobedo]

Nice. Thanks to both of you. I know exactly what to do.