Answer:
Mass of the sample = m = 894.0 g
Initial temperature of the sample = T_i=-5.8^oC=267.35 KT
i
=−5.8
o
C=267.35K
Final temperature of the sample = T_f=17.5^oC=290.65 KT
f
=17.5
o
C=290.65K
Change in temperature = \Delta T=(T_f-T_i)ΔT=(T
f
−T
i
)
Specific heat capacity of the substance = c
Heat required to raise the temperature of a 894.0g sample = Q
Q = 4.90kJ = 4900 J
Q=m\times c\times \Delta T=m\times c\times (T_f-T_i)Q=m×c×ΔT=m×c×(T
f
−T
i
)
4900 J=894.0 g\times c\times (290.65 K-267.35 K)4900J=894.0g×c×(290.65K−267.35K)
c=0.23523 J/g K\approx 0.235 J/g Kc=0.23523J/gK≈0.235J/gK
The chemist will report the specific heat capacity of the substance as 0.235 J/g K.
