Question

Question 1 A 250-g sample of a metal absorbed 27.0 J of heat as it’s temperature increased 10.0•C to 15•C. What is the specific heat of the metal ? Question 2 The temperature of a sample of water increases from 15.0•C to 30•C as it absorbes 5000 J of heat. What is the mass of the sample? Specific heat of water is 2.03 J/g•C

Answer 1

For the first question, we will use the calorimetry formula, which is:

Q = mcΔT

We have:

Q = is heat, 27.0 J

m = is mass, 250.0 g

ΔT = is the change in temperature, which will be, Final T - Initial T (15 - 10 = 5°C)

Now we add these values to the formula:

27.0 J = 250.0 g * c * 5°C

27.0 = 1250c

c = 0.0216 J/g°C, this is the specific heat for this metal

For the second question, we will again use the same calorimetry formula, but now we are going to look for the mass:

Q = mcΔT

We have:

Q = 5000 J

m = ?

c = 2.03 J/g°C

ΔT = 30 - 15 = 15°C

Now we add these values to the formula:

5000 J = m * 2.03 J/g°C * 15°C

5000 J = 30.45m

m = 164.2 grams, this is the mass for this question