Question

Table 1: Mass Data Mass (g) Trial 1 Trial 2 Trial 3 Water 49.7 50.6 50.2 Metal Strip 17.9 18 18 Table 2: Specific Heat Data
Time (minutes)	Temperature (°C)
	Trial 1	Trial 2	Trial 3
Initial	23.9	23.5	25.4
5 minutes	25.7	25.5	27.6
6 minutes	25.7	25.4	27.6
7 minutes	25.6	25.4	27.5
8 minutes	25.6	25.4	27.5
9 minutes	25.6	25.4	27.4
10 minutes	25.5	25.4	27.4
T	1.6	2.0	2.0

Based on the data in the tables above, Calculate the specific heat of the metal strip for each trial here. What is the average specific heat?

Answer 1

Solution :-

The heat given by metal is absorbed by the water therefore we can set up the equation to calculate the specific heat of metal

Initial temperature of the metal is 100 C

Lets calculate the specific heat of metal for trial 1

Equation

-q metal = q water

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

Where, m=mass, s= specific heat , delta T= temperature change

- 17.9 g*s*(25.6C-100C) = 49.7 g* 4.184 J per gC * 1.6 C

1466 g C *s = 333 J

S= 333 J / 1466 gC

S=0.250 J/gC

Now lets calculate the specific heat for trial 2

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

- 18 g*s*(25.4C-100C) = 50.6 g* 4.184 J per gC * 2.0 C

1341 g C *s = 423.4 J

S= 423.4 J / 1341 g C

S= 0.316 J/gC

Now lets calculate the specific heat for trial 3

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

- 18 g*s*(27.6C-100C) = 50.2 g* 4.184 J per gC * 2.0 C

1303.2 g C * s = 420.1 J

S= 420.1 J / 1303.2 g C

S=0.322 J/gC

Now lets calculate the average specific heat

Average specific heat = [0.250 +0.316+0.322] / 3

                                         = 0.296 J/gC