Question

Table 1: Mass Data
	Mass (g)
	Trial 1	Trial 2	Trial 3
Water	48.0 g	48.5 g	48.3 g
Metal Strip	36.9 g	37.0 g	37.2 g

Table 2: Specific Heat Data
Time (minutes)	Temperature (°C)
	Trial 1	Trial 2	Trial 3
Initial	35.1 C	35.6 C	35.6 C
5 minutes	35.3 C	36.4 C	35.8 C
6 minutes	35.0 C	36.0 C	35.4 C
7 minutes	34.6 C	35.7 C	35.4 C
8 minutes	34.6 C	35.5 C	35.3 C
9 minutes	34.3 C	35.2 C	35.1 C
10 minutes	34.2 C	35.0	34.8 C
DT

1. Average Specific Heat Capacity of the Unknown Metal:

2. Calculate the specific heat of the metal strip for each trial here. What is the average specific heat?

temperature of water is in table 2. temperature was taken then monitiored over time increments. initial is starting temp.

Remember the initial temperature of the metal is 100 C, and the final temperature is the same at the water in the calorimeter.

Answer 1

Solution :-

The highest temperature recorded in the time interval is considered as the final temperaure of mixture

Trial 1

Mass of water =48.0 g

Mass of metal strip = 36.9 g

Initial temperature of water = 35.1 C

Initial temperature of metal strip = 100 C

Final temperature = 35.3 C

Heat lost by metal is same as heat absorbed by water

-q metal = q water

-m*s*delta T = m*s*delta T

where, m= mass , s= specific heat , delta T= change in temperautre

So lets put the values in the formula and solve for the specific heat of metal in trial 1

-m*s*delta T = m*s*delta T

-36.9 g* s * (35.3 C – 100 C) = 48.0 g * 4.184 J per g C *(35.3 C – 35.1 C)

2387.43 *s = 40.1664

S= 40.1664 / 2387.43

S= 0.0168 J/g C

Therefore specific heat of metal for trial 1 = 0.0168 J/gC

Trial 2

-m*s*delta T = m*s*delta T

-37.0 g* s * (36.4 C – 100 C) = 48.5 g * 4.184 J per g C *(36.4 C – 35.6 C)

2353.2*s = 162.3392

S= 162.3392 / 2353.2

S= 0.0690 J/gC

Therefore specific heat of metal in trial 2 = 0.0690 J/gC

Trial 3

-m*s*delta T = m*s*delta T

-37.2 g* s * (35.8 C – 100 C) = 48.3 g * 4.184 J per g C *(35.8 C – 35.6 C)

2388.24*s =40.42

S=40.42 / 2388.24

S= 0.0169 J/gC

Therefore specific heat of metal in trial 3 = 0.0169 J/gC

Average specific heat = [trial 1 + trial 2 + trial 3]/3

                                      =[0.0168+0.0690+0.0169]/3

                                      = 0.0342 J/gC