Question

If 1.0 kJ of heat was added to each 50.0 g sample, which would increase temperature by the greatest amount? I can't seem to figure out the correct answer.

If 1.0 kJ of heat was added to each 50.0 g sample, which would increase temperature by the greatest amount? O aluminum (0.895 J/(g:°C)) lead (0.129 J/(g:°C)) O copper (0.377 J/(g:°C)) O iron (0.448 J/(g°C))

Answer 1

Answer: lead(0.129 J/ (g.°C) -----> option(b)

Step 1: Explanation

We know

Q=m×c×ΔT

where, m = mass of the material, c =  specific heat capacity of the material, and ΔT = temperature change

Step 2: Calculation

m = 50 g

Q = 1 kJ = 1000 J [ since 1 kJ = 1000 J ]

We know

Q=m×c×ΔT

or, => ΔT= Q / m×c

since here mass and Q is constant for all the given samples

hence, now increase in temperature is depend on specific heat constant

temperature will higher at lower specific heat constant because   ΔT is inversely proportional to c

Case 1: for aluminum ,   ΔT= 1000 J / 50 g × 0.895 J/(g.°C) = 22.35 °C

Case 1: for lead ,   ΔT= 1000 J / 50 g × 0.129 J/(g.°C) = 155 °C

Case 1: for copper  ,   ΔT= 1000 J / 50 g × 0.377 J/(g.°C) = 53 °C

Case 1: for iron ,   ΔT= 1000 J / 50 g × 0.448 J/(g.°C) = 44.6 °C

So we can see that as C decreases change in temperature increases and higher is for lead