Question

In a lab experiment involving calorimetry with brass and lead samples, a perfect calorimeter with no heat loss in which any heat lost from the metal was gained by the water was assumed. If the calorimeter was not perfect and some heat was lost to the environment, would this cause your calculated specific heat capacity of a solid (c) values to be higher or lower than the actual specific heat capacity of a solid (c) values?

Answer 1

Heat lost by metal = m.s.∆T

Heat gained by water = m1.s1.∆T1

∆T1 would be smaller than actual value and ∆T would be larger than the actual value .

m.s.∆T = m1.s1.∆T1 + (heat lost to environment)

s = m1s1∆T1/(m.∆T)

The specific heat capacity of solid would be lower than the actual specific heat capacity is solid .