# Specific Heat Capacity of water in a kettle

1. Jan 30, 2015

### Dave Jones

1. The problem statement, all variables and given/known data

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)

2. Relevant equations

3. 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

2. Jan 30, 2015

### Quantum Defect

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. Jan 30, 2015

### Staff: Mentor

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. Jan 30, 2015

### Dave Jones

Oh..But i thought you always take the mass in SI which is kg?

Ok thanks

5. Jan 30, 2015

### Dave Jones

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. Jan 30, 2015

### Staff: Mentor

Right. Clearly incorrect. Hint: What are the units of the 322 value you've used?

7. Jan 30, 2015

### Staff: Mentor

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. Jan 30, 2015

### Dave Jones

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. Jan 30, 2015

### Dave Jones

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

Been a long day...lol.