Question

A 15.0g sample of an unknown metal at 85.0*C was placed in 60.0g of water at 20.0*C. the final temperature of the metal and water was 25.8*C. calculate the metal's specific heat (please show the work)

how much heat in calories and joules is required to raise the temperature of 75.0g of a substance (specific heat is 1.80cal/g x *C) from 25.0*C to 95.0*C? please show the work. thank you

Answer 1

Ans. #1. Amount of gained by water when its temperature increases from 20.0⁰C to 25.8⁰C is given by-

            q = m s dT                            - equation 1

Where,

q = heat absorbed

m = amount of water in moles

s = specific heat of water [ 4.184 J g^-10C^-1]

dT = Final temperature – Initial temperature

= 25.8⁰C – (20.0⁰C) = 5.80⁰C

Putting the values in equation 1-

            q = 60.0 g x (4.184 J g^-10C^-1) x 5.8⁰C = 1456.032 J

# Heat gained by water must be equal to heat lost by sample when it cools from 85.0⁰C to 25.8⁰C (equilibrium temperature).

            dT for sample = 85.0⁰C – 25.8⁰C = 59.2⁰C

            q for sample = 1456.032 J            

            m of sample = 15.0 g

Putting the values in equation 1 for the sample-

            1456.032 J = 15.0 g x s x 59.2⁰C

            Or, s = 1456.032 J / ( 888 g ⁰C) = 1.64 J g^-10C^-1

Hence, specific heat of sample = 1.64 J g^-10C^-1

#2. dT = 95.0⁰C – 25.0⁰C = 70.0⁰C

            m = 75.0 g

            s = 1.80 cal g^-10C^-1

Putting the values in equation 1-

            q = 75.0 g x (1.80 cal g^-10C^-1) x 70.0⁰C = 9450.0 cal

Hence, required amount of heat = 9450.0 cal                                        ; [1 cal = 4.184 J]

                                                = 39538.8 J