Question

A 248-g sample of copper is dropped into 390 g of water at 22.6oC. The final temperature was measured to be 39.9 oC. Calculate the initial temperature of the copper.
Density of water 1.00g/mL. Specific heat of copper = .384 J/g oC

Answer 1

Given : 1) mass of water = 390 g

2) mass of Copper = 248 gm

3) initial temp. of water = 22.6 ⁰ C

4) initial temp. of Cu = ?

5) Final temp of both = 39.9 ⁰ C

6) Specific heat of water C _water = 4.184 J/ g ⁰ C

7) Specific heat of copper = 0.384 J/ g ⁰ C

In this case, heat given by metal is taken by water .Therefore we can write q _Cu + q _H2O = 0

We know that, Heat absorbed or emitted by any substance is given as q = m x C x (T final - T initial ) (1)

Where q is a heat absorbed or emitted, m is a mass of a body, C is a specific heat capacity of a body, T is a temperature of a body.

$\therefore$ [ m $\times$ C $\times$ ( T final - T initial) ] _copper + [ m $\times$ C $\times$ ( T final - T initial) ] _water = 0

[ 248 g $\times$ 0.384 J/ g ⁰ C $\times$ ( 39.9 ⁰ C - T initial) ] _copper + [ 390 g $\times$ 4.184 J/ g ⁰ C $\times$ ( 39.9 ⁰ C - 22.6 ⁰ C) ] _water = 0

[ 248 g $\times$ 0.384 J/ g ⁰ C $\times$ ( 39.9 ⁰ C - T initial) ] _copper + [ 390 g $\times$ 4.184 J/ g ⁰ C $\times$ 17.3 ⁰ C] _water = 0

[ 248 g $\times$ 0.384 J/ g ⁰ C $\times$ ( 39.9 ⁰ C - T initial) ] _copper + 28229.4 J = 0

[95.232 J/  ⁰ C $\times$ ( 39.9 ⁰ C - T initial) ] _copper + 28229.4 J = 0

[95.232 J/  ⁰ C $\times$ ( 39.9 ⁰ C - T initial) ] _copper = - 28229.4 J

( 39.9 ⁰ C - T initial) ] _copper = - 28229.4 J / 95.232 J  ⁰ C = - 296.4 ⁰ C

( 39.9 ⁰ C - T initial) ] _copper = - 296.4 ⁰ C

[T initial) ] _copper = 39.9 ⁰ C + 296.4 ⁰ C = 336.3 ⁰ C

ANSWER : Initial temperature of Copper metal = 336.3 ⁰ C