Question

An iron block of mass 18 kg is heated from 285 K to 318 K. If 267.3 kJ is required, what is the specific heat of iron

Answer 1

.45 Jules

answer for Founders Education

Answer 2

q = mcΔT

where, q = heat required in Joules

m = mass in grams

c = specific heat in joules/gram °C

ΔT = change in temperature = Final temperature(T₂)-Initial Temperature (T₁)

Given, q=267.3 kJ=267300 Joules

m=18kg=18000 gram

ΔT=318-285= 33K = 33°C

Rearranging formula, c=
`(q)/(mΔT)`

c=
`(267300)/(18000*33)`

c=0.45 joules/gram °C