Respuesta :
Answer:
The chemist will report the specific heat capacity of the substance as 0.235 J/g K.
Explanation:
Mass of the sample = m = 894.0 g
Initial temperature of the sample = [tex]T_i=-5.8^oC=267.35 K[/tex]
Final temperature of the sample = [tex]T_f=17.5^oC=290.65 K[/tex]
Change in temperature = [tex]\Delta T=(T_f-T_i)[/tex]
Specific heat capacity of the substance = c
Heat required to raise the temperature of a 894.0g sample = Q
Q = 4.90kJ = 4900 J
[tex]Q=m\times c\times \Delta T=m\times c\times (T_f-T_i)[/tex]
[tex]4900 J=894.0 g\times c\times (290.65 K-267.35 K)[/tex]
[tex]c=0.23523 J/g K\approx 0.235 J/g K[/tex]
The chemist will report the specific heat capacity of the substance as 0.235 J/g K.
An 894.0 g sample with a specific heat capacity of 2.35 × 10⁻⁴ kJ/g.° C, increases its temperature from -5.8 °C to 17.5 °C when absorbing 4.90 kJ of heat.
A chemist has a sample with a mass of 894.0 g. When it absorbs 4.90 kJ of heat its temperature increases from -5.8 °C to 17.5 °C. The chemist can calculate the specific heat capacity of the substance using the following expression.
[tex]Q = c \times m \times \Delta T[/tex]
where,
- c: specific heat capacity of the substance
- m: mass of the sample
- ΔT: change in the temperature of the sample
[tex]Q = c \times m \times \Delta T\\\\c = \frac{Q}{m \times \Delta T} = \frac{4.90 kJ}{894.0 g \times (17.5 \° C - (-5.8 \° C))} = 2.35 \times 10^{-4} kJ/g\° C[/tex]
An 894.0 g sample with a specific heat capacity of 2.35 × 10⁻⁴ kJ/g.° C, increases its temperature from -5.8 °C to 17.5 °C when absorbing 4.90 kJ of heat.
Learn more: https://brainly.com/question/11194034
