Question

When a substance having mass 3kg receives 600cal of heat, it's temperature increases by 10degree celcius. What is the specific heat of the substance.

Answer 1

∆Q = mc∆t

Here, ∆Q = 600 cal

m = 3 kg

c = ?

∆t = 10

So, 600 = 3×c×10

c = 600/30

c = 20 Cal/kg °C.