Specific Heat Capacity

Round your answer to 4 decimal places.

The specific heat capacity of water is 4.2 kJ/kg°C

Answer explanation

Rearrange ΔQ = m x c x ΔT to calculate for mass.

m = ΔQ / c x ΔT

m = 8.36 kJ / 4.2 kJ/kg°C x 40°C

m = 0.049761904 kg

m = 0.0498 kg (4dp)

Tags

NGSS.HS-PS3-1

NGSS.HS-PS3-4

A pupil wanted to investigate what the specific heat capacity of orange juice was.

They measured the temperature of their 250 mL glass of orange juice and recorded the temperature as 25 °C and then began heating the glass. When they stopped recording at 35 °C, 9.73 kJ of energy had been absorbed by the orange juice.

Use this information to calculate the specific heat capacity of orange juice.

Answer explanation

C = E_h / mΔT

 

C= 9.73 kJ / 0.25 kg * 10 °C

 

C = 3.89 kJ kg^-1‑ °C^-1

Tags

NGSS.HS-PS3-1

NGSS.HS-PS3-4

A scientist conducted an experiment to investigate how much heat energy in kJ would be required to heat 500 mL of methanol by 30 °C. The specific heat capacity of methanol is 2.53 kJ kg⁻¹ °C⁻¹.

Use this information to calculate how much heat energy would be required.

Answer explanation

E_h = cmΔT

 

E_h = 2.53 kJ kg⁻¹ °C⁻¹ x 0.5 kg x 30 °C

 

E_h = 37.95 kJ

Tags

NGSS.HS-PS3-1

100 mL of water was heated with 60 kJ of heat energy. What temperature change would this have caused?

Specific heat capacity of water is ~4.2 kJ/kg°C

Round your answer to 3 significant figures.

Answer explanation

ΔT = E_h / cm

ΔT = 60 kJ/ 4.2 kJ/kg°C x 0.1 kg

ΔT = 142.8571

ΔT = 143°C (3sf)

Tags

NGSS.HS-PS3-1

30g of water was heated and its temperature was raised by 12°C. Calculate how much heat energy was supplied to the water during this time.

The specific heat capacity of water is 4200 J/kg°C

Answer explanation

ΔQ = m x c x ΔT

ΔQ = 0.03 kg x 4200 J/kg°C x 12 °C

ΔQ = 1512 J

Tags

NGSS.HS-PS3-1

NGSS.HS-PS3-4

The specific heat capacity of marble is 832 J/kg°C. If a 700 gram block of marble was heated by 34°C, how much heat energy would have been required to do this?

Do not round your answer.

Answer explanation

ΔQ = m x c x ΔT

ΔQ = 0.7 kg x 832 J/kg°C x 34°C

ΔQ = 19801.6 J

Tags

NGSS.HS-PS3-1

A typical nuclear bomb requires ~15 grams of uranium to be operational. Calculate the change in temperature at the centre of a nuclear explosion if the bomb provided 172, 500 kJ of energy to the uranium and the specific heat capacity of uranium is 115 J/kg/°C.

Do not include commas in your answer

Answer explanation

ΔQ = m x c x ΔT

ΔT = ΔQ / cm

ΔT = 172500000 J (convert kJ to J here) / 115 x 0.015 kg

ΔT = 100,000,000 °C

So the temperature at the centre of a nuclear explosion is 100 million degrees Celsius.

[For comparison, the surface of the sun is ~5800]

Tags

NGSS.HS-PS1-8

NGSS.HS-PS3-4

A student set up an experiment similar to the one shown to investigate the specific heat capacity of a 2500 gram block of rubber. The block was supplied with 250.625 kJ of energy which caused the temperature of the block to rise by 50 °C.

Calculate the specific heat capacity of rubber in J/kg°C.

Answer explanation

ΔQ = m x c x ΔT

c = ΔQ / mΔT

Question asks for value of c in J/kg°C so convert the kJ into J (250625 J).

c = 250625 J / 2.5 kg x 50°C

c = 2005 J/kg°C

Tags

NGSS.HS-PS3-1

NGSS.HS-PS3-4

A student set up an experiment similar to the one shown to investigate the specific heat capacity of a block of rubber. The block was supplied with 104356 J of energy which caused the temperature of the block to rise by 27 °C. The specific heat capacity of rubber is 2005 J/kg°C.

Use this information to calculate the mass of the block of rubber. Give your answer to two decimal places.

Answer explanation

ΔQ = m x c x ΔT

m = ΔQ / cΔT

c = 104356J / 2005 J/kg°C x 27°C

m = 1.927699 kg

m = 1.93 kg (2dp)

Tags

NGSS.HS-PS3-1

NGSS.HS-PS3-4

If the sun provided 15000 J of heat energy to a 1kg block of concrete, how much would the temperature of the block increase by? (ignoring loss of heat to the environment). The specific heat capacity of concrete is 829 J/kg°C.

Give your answer to two decimal places.

Answer explanation

ΔQ = m x c x ΔT

ΔT = ΔQ / mc

ΔT = 15000 J / 1 kg x 829 J/kg°C

ΔT = 18.09408

ΔT = 18.09 °C (2dp)

Tags

NGSS.HS-PS3-1