Question

PLEASE HELP!!

Answer 1

As a 15.1-gram sample of a metal absorbs 48.75 J of heat, its temperature increases 25.0K. $0.129 \frac{\mathrm{J}}{\mathrm{g} \mathrm{K}}$ is the specific heat capacity of the metal

Answer: Option A

Explanation:

Specific heat term explains the amount of heat needs to be added with unit mass in order to increase the temperature by a degree Celsius. Its formula is given by,

              $Q=m \times c \times \Delta T$

Where,

$\Delta T=\text { final } T-\text {initial} T=25.0 \mathrm{K}$

$Q \text { is the heat energy in Joules }=48.75 \text { Joules }$

$c \text { is the specific heat capacity }$

$m \text { is the mass of the metal }$

Plugging in the values  

$c=\frac{Q}{m \times \Delta T}=\frac{48.75 \mathrm{J}}{15.1 \mathrm{g} \times 25.0 \mathrm{K}}=\frac{48.75 \mathrm{J}}{377.5 \mathrm{g} \mathrm{K}}=0.129 \frac{\mathrm{J}}{\mathrm{g} \mathrm{K}}(\text { Answer })$

Answer 2

Answer is: the specific heat capacity of the metal is A) 0.129 J/gK.
m(metal) = 15,1 g.
Q = 48,75 J.
ΔT = 25 K.
Q = C · ΔT · m(metal).
C = Q ÷ ΔT · m(metal).
C = 48,75 J ÷ 25 K · 15,1 g.
C = 0,129 J/g·K.