A technician heats a solid tile and records these readings.
| Quantity | Reading |
|---|---|
| Mass of tile | 0.60 kg |
| Starting temperature | 20 ∘C20\ ^\circ\mathrm{C}20 ∘C |
| Final temperature | 40 ∘C40\ ^\circ\mathrm{C}40 ∘C |
| Energy transferred to tile | 10800 J |
The technician uses ΔQ=mcΔθ\Delta Q=mc\Delta\thetaΔQ=mcΔθ to find the specific heat capacity of the tile. Which value does the technician obtain?
450 J kg−1 ∘C−1450\ \mathrm{J\,kg^{-1}\,{}^\circ C^{-1}}450 Jkg−1∘C−1
540 J kg−1 ∘C−1540\ \mathrm{J\,kg^{-1}\,{}^\circ C^{-1}}540 Jkg−1∘C−1
900 J kg−1 ∘C−1900\ \mathrm{J\,kg^{-1}\,{}^\circ C^{-1}}900 Jkg−1∘C−1
18000 J kg−1 ∘C−118000\ \mathrm{J\,kg^{-1}\,{}^\circ C^{-1}}18000 Jkg−1∘C−1
110 exam-style questions on Edexcel GCSE Physics 14.2 Specific heat capacity and latent heat, covering 14.2.1 Heating, temperature and changes of state, 14.2.2 Specific heat capacity and specific latent heat, 14.2.3 Specific heat capacity calculation, 14.2.4 Specific latent heat calculation, and 14.2.5 Reducing energy transfer through thermal insulation. Each one has a worked solution and a mark scheme showing where the marks go.