Diagram shows the heating curve for 0.5 kg heated by a
100 W immersion heater. What
is the specific heat capacity of
the liquid?

35

Explanation:

Given the following :

M = 0.5 kg

Power of heater = 100W

From the graph ;

Time (t) = 7seconds

Change in temperature(Dt) = t2 - t1

Where t2 = 70°C, t1 = 30°C

Dt = (70 - 30)°C = 40°C

Recall Q = iVt

IV = power

Q = mcDt

c = specific heat capacity of liquid

mcDt = ivt

mcDt = power × t

0.5 × c × 40°C = 100 × 7

20°C × c = 700 J

c = 700 J/ 20°C

c = 35 J / kg°C

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

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