Find the temperature of Liquid A after 12 minutes

[Richie Smash]

Homework Statement

A liquid A initially at temperature T1= 28.4°C, is heated by an immersion heater in container. The resulting temperature T2 is recorded at 1 minute intervals and the following results obtained.
(I will post a picture)
(a) Complete the table by computing temperature changes ΔT=T2-T1
I've done this and the missing values are: 7.6, 14.7, 24.1, 32.1,40.6,46.7,55.9 degrees Celsius.
(b) Plot a graph of ΔT against time (t). I've also done this.
(c) Find the slope S of the graph, I've also done this and I got 8.33.
(d) Cp the Specific Heat capacity of the liquid is related to the slope S by : Cp=2000/S J Kg^-1K^-1.
Find the Specific Heat Capacity of the Liquid.
I've also done this and got 240.1J Kg^-1K^-1.
(e) What would be the temperature of the liquid after 12 minutes?
This is where I'm stuck.

Homework Equations

Eh=m*C*ΔT

The Attempt at a Solution

I'm really stuck, I was thinking now I have the specific heat capacity i could find the energy supplied, then relate that to the temperature change but I don't have the mass. They basically want X°C-28.4°C and that would give you the temperature change after 12 minutes then simply add to the initial temperature to get the final.

But I'm pretty unsure I've been racking my brain looking for formulas linking temperature and time.

 

[TSny]

(e) What would be the temperature of the liquid after 12 minutes?
This is where I'm stuck.

In part (c) you didn't include the units for the slope. This should help with answering part (e). Interpret what the slope is telling you.

 

[Richie Smash]

The unit would then be degrees Celsius per minute?

 

[TSny]

The unit would then be degrees Celsius per minute?

Yes.

 

[Richie Smash]

Wait, so it's as simple as doing 8.33 degrees celsius per min *12 +28.4 to give me 128.36°C

 

[TSny]

Wait, so it's as simple as doing 8.33 degrees celsius per min *12 +28.4 to give me 128.36°C

Yes, that looks right. However, I don't know how you got your value of 8.33 C^o/min for the slope. It appears to be a little too large.

 

[Richie Smash]

On My graph I drew a line of best fit, so I suppose there was human error involved in finding the gradient.

 

[TSny]

On My graph I drew a line of best fit, so I suppose there was human error involved in finding the gradient.

Ok. I think your work looks good.

 

[Daynea G]

How did the graph look?

 

[haruspex]

How did the graph look?

What graph?

 

[Daynea G]

The graph for the work

 

[haruspex]

The graph for the work

I only see one image attached in the thread, the image in post #1. To the right of the image I see part of a sheet of graph paper, but no actual graph.

 

[Daynea G]

Me too that's why I asked

 

[kuruman]

There seems to be an ambiguity as to the entries in the row labeled "Temperature changes ΔT/^oC". The first entry is obviously (ΔT)₁ = T₂ - T₁. What about the ith entry? Is it (ΔT)_i = T_i+1 - T_i or, as OP surmised, (ΔT)_i = T_i+1 - T₁? The former set of "instantaneous slope" entries is probably more useful because the average gives an approximate value of the slope that could be verified with a linear fit.