Question

A 23.50 g sample of a substance is initially at 24.9 °C. After absorbing 2495 J of heat, the temperature of the substance is 144.0 °C. What is the specific heat (c) of the substance?

Answer 1

mass of substance (m) =23.5 g

Change in temperature ($\Delta$T) = 144 - 24.9 = 119.1 0C

Absorbed heat (Q) = 2495 J,

we have to find, specific heat (c) = ?

Since we know, Q= c.m.$\Delta$T

Rearranging, c = Q/ [m.$\Delta$T] = 2495 / [23.5 * 119.1]

c = 0.89