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