Explanation
o solve this , we need to use the formula:
[tex]\begin{gathered} Q=m*c*(T_f-T_0) \\ where \\ m\text{ is the mass measured in kg} \\ c\text{ is the specific heat of silver=0.129 J/\degree C} \\ T_f\text{ and T}_o\text{ are the final and initial temperatures} \end{gathered}[/tex]so
Step 1
) Let
[tex]mass=31\text{ gr}[/tex]so, we need to convert from grams to kilograms ,to do that, divide the amount in grams by 1000
so
[tex]mass_{kg}=\frac{31}{1000}=0.031\text{ kg}[/tex]and
[tex]\begin{gathered} T_f=\text{1\degree C} \\ T_0=0°\text{ C} \end{gathered}[/tex]b) now, replace in the formula
[tex]\begin{gathered} Q=mc(T_{f}-T_{0}) \\ Q=0.031gr*0.129\frac{J}{g°C}(1°C-0°C) \\ Q=0.003999\text{ Joules} \end{gathered}[/tex]so, the answer is 0.003999 Joules
hope this helps you