Question

In an experiment, 26.0 g of metal was heated to 98.0°C and then quickly transferred to 150.0 g of water in a calorimeter. The initial temperature of the water was 20.5°C, and the final temperature after the addition of the metal was 32.5°C. Assume the calorimeter behaves ideally and does not absorb or release heat.

What is the value of the specific heat capacity (in J/g•°C) of the metal?

Answer 1

Given: 1) mass of water = 150.0 g

2) Mass of metal = 26.0 g

3) Initial temp. of water = 21.5 ⁰C

4) Initial temp. Of metal = 98.0 ⁰C

5) Final temp of both = 32.5 ⁰C

6) Specific heat of water C _water = 4.184 J/g⁰C

In this case, heat given by metal is taken by water .Therefore we can write

                                           q _Metal = - q _water

We know that heat is given as q= C x m x T

                                      Where q= heat given or absorbed

                                                 C= specific heat capacity of material

                                             T = temp difference during heat change  = T _final- T _initial

Therefore, C _metal x m _metal x (T _final- T _initial) _metal = - C _water x m _water x (T _final- T _initial) _water

C _metal x 26.0 g x (32.5 ⁰ C - 98.0 ⁰ C) _metal = - 4.184 J/g ⁰ C  x 150.0 g x ( 32.5 ⁰  C - 20.5 ⁰ C) _water

                           C _metal x - 1703.0 = - 7531.2

                                           C _metal =- 7531.2 / - 1703.0

                                            C _metal = 4.422 J/ g ⁰ C

ANSWER : Specific heat capacity of metal = 4.422 J/ g ⁰ C