A 237-g piece of wire is heated, and the temperature of the wire changes from 25*C to 107*C. The amount of heat absorbed is 722 cal. What is the specific heat of the wire? Show all work, including the equation, and include units

The answer to your question is  C = 0.037 cal/g°C

Explanation:

Data

mass of the wire = m = 237 g

temperature 1 = T1 = 25°C

temperature 2 = T2 = 107°C

Heat = Q = 722 cal

Specific heat = C

Process

1.- Write the formula to find the specific heat

            Q = mC(T2 - T1)

-Solve for C

            C = Q / m(T2 - T1)

2.- Substitution

             C = 722 / 237(107 - 25)

3.- Simplification

              C = 722 / 237(82)

              C = 722 / 19434

4.- Result

               C = 0.037 cal/g°C