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.

 

[Nickie]

Dear scientist,

Thank you for reaching out for help with this problem. Let's break it down step by step to find the correct solution.

First, we know that to increase the temperature of the vessel by 1 K, 50 J of energy must be supplied. This means that the specific heat capacity of the vessel is 50 J/(kg*K). However, since the mass of the vessel is unknown, we can't use this information to calculate the specific heat capacity of the liquid.

Next, we know that 250 g of liquid is added to the vessel. This means that the mass of the liquid is 0.25 kg.

We also know that the heating spiral develops 15 W of power. This means that for every second, 15 J of energy is supplied to the vessel. We can use this information to calculate the change in temperature of the liquid in one second.

ΔT = (15 J/s)/(0.25 kg * 3.0 kJ/(kg*K)) = 20 K/s

This means that the temperature of the liquid will increase by 20 K every second until it reaches a stable temperature.

Now, we also know that the temperature decreases by 1.2 K per minute once the heat supply is turned off. This means that the rate of temperature change is -0.02 K/s.

Since we want to find the specific heat capacity of the liquid, we can use the formula c = Q/(m * ΔT) where Q is the energy supplied, m is the mass of the liquid, and ΔT is the change in temperature.

c = (15 J/s)/(0.25 kg * 20 K/s) = 3.0 kJ/(kg*K)

This is the same result you got in your attempt, but we can see that it is not the correct answer according to the textbook. The reason for this is that we need to take into account the heat loss from the vessel itself.

To do this, we can use the equation E = mcΔT where m is the mass of the vessel, c is the specific heat capacity of the vessel, and ΔT is the change in temperature.

Since we know that the vessel loses heat at a rate of 1.2 K per minute, this means that the energy lost per second is (1.2 K/min)/(60 s/min) = 0.02 K/s.

Substituting this into the equation,