Specific Heat Capacity of water in a kettle

  • #1

Homework Statement



mass, m, of water in kettle

1.5 kg

power rating, P, of kettle

2.1 kW

time interval, t, for heating

322 seconds

starting temperature

4 °C = 277 K

finishing temperature

100 °C = 373 K

temperature change, ΔT

96 K

electrical energy supplied, E = P × t

(2.1x10^3)(322) = 6.8x10^5 J

calculate the specific heat capacity:


c = q/(m)(delta t)



Homework Equations




The Attempt at a Solution



Why do i keep getting the wrong answer?

q = 6.8 x 10^5
m = 1.5kg
t = 322

Thus c = 6.8 x 10^5/1.5*322 = 1407.8674948240165631469979296066252587991718426501035

But im sure that's wrong, it's not supposed to be the exact figure, as this was a rough experiment done at home. It's way off, where am i going wrong? thanks
 

Answers and Replies

  • #2
Quantum Defect
Homework Helper
Gold Member
495
116

Homework Statement



mass, m, of water in kettle

1.5 kg

power rating, P, of kettle

2.1 kW

time interval, t, for heating

322 seconds

starting temperature

4 °C = 277 K

finishing temperature

100 °C = 373 K

temperature change, ΔT

96 K

electrical energy supplied, E = P × t

(2.1x10^3)(322) = 6.8x10^5 J

calculate the specific heat capacity:


c = q/(m)(delta t)



Homework Equations




The Attempt at a Solution



Why do i keep getting the wrong answer?

q = 6.8 x 10^5
m = 1.5kg
t = 322

Thus c = 6.8 x 10^5/1.5*322 = 1407.8674948240165631469979296066252587991718426501035

But im sure that's wrong, it's not supposed to be the exact figure, as this was a rough experiment done at home. It's way off, where am i going wrong? thanks
A couple of things. The answer should have units of J g^-1 K^-1 (i.e. J/g K)

You took Joules and devided by mass (kg) and time (sec). Your answer has units of J/kg s -- not right.

You want to take the heat (which you have) divide by the mass in g and divide by the change in T.
 
  • #3
gneill
Mentor
20,913
2,861
Thus c = 6.8 x 10^5/1.5*322 = 1407.8....
Largely your problem is calculator operator error. You've divided by 1.5 then multiplied by 322 when the 322 should be in the denominator. Parentheses can help manage the order of operations.
 
  • #4
A couple of things. The answer should have units of J g^-1 K^-1 (i.e. J/g K)

You took Joules and devided by mass (kg) and time (sec). Your answer has units of J/kg s -- not right.

You want to take the heat (which you have) divide by the mass in g and divide by the change in T.
Oh..But i thought you always take the mass in SI which is kg?

Largely your problem is calculator operator error. You've divided by 1.5 then multiplied by 322 when the 322 should be in the denominator. Parentheses can help manage the order of operations.
Ok thanks
 
  • #5
This still isn't making sense to me:

Right so the equation for specific heat is:

c = q/(m*delta t) right?

So if we are using 6.8 x 10^ 5 as q, and 1500grams for mass, and change in t is 322 seconds, that would give us:

c = (6.8 x 10^5)/((1500)(322))

But that gives me an answer of like 1.4 J/kg s
 
  • #6
gneill
Mentor
20,913
2,861
c = (6.8 x 10^5)/((1500)(322))

But that gives me an answer of like 1.4 J/kg s
Right. Clearly incorrect. Hint: What are the units of the 322 value you've used?
 
  • #7
20,844
4,543
This still isn't making sense to me:

Right so the equation for specific heat is:

c = q/(m*delta t) right?

So if we are using 6.8 x 10^ 5 as q, and 1500grams for mass, and change in t is 322 seconds, that would give us:

c = (6.8 x 10^5)/((1500)(322))

But that gives me an answer of like 1.4 J/kg s
The symbol t does not always refer to time (322 sec). Sometimes it refers to temperature difference (96 C). Which do you think it refers to in this context?

Chet
 
  • #8
Right. Clearly incorrect. Hint: What are the units of the 322 value you've used?
Seconds, aren't i supposed to be using that? This is confusing lol.

I always get told you're supposed to use SI units, and for mass that's kg and time s :(
 
  • #9
The symbol t does not always refer to time (322 sec). Sometimes it refers to temperature difference (96 C). Which do you think it refers to in this context?

Chet
Oh wow im stupid haha. I can't believe i've just done that.

Been a long day...lol.
 

Related Threads on Specific Heat Capacity of water in a kettle

  • Last Post
Replies
16
Views
2K
Replies
6
Views
2K
Replies
7
Views
2K
Replies
2
Views
5K
  • Last Post
Replies
2
Views
15K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
7
Views
824
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
9
Views
3K
Top