Question

Use this equation to calculate the specific heat capacity of each metal:

cmetal =–StartFraction c subscript water times m subscript water times delta t subscript water over m subscript metal times delta t subscript metal EndFraction.
cwater = sp. heat capacity of water (4.18 J/gC)

mwater = mass of the water in the cup

Twater = change in water temperature

mmetal = mass of the metal

Tmetal = change in metal temperature



Record these values in Table B of your Student Guide.

What is the specific heat for the aluminum wire?
J/gC

What is the specific heat for the steel wire?
J/gC

What is the specific heat for the lead pellets?
J/gC

Answer 1

I can calculate those, but I need the measured values from your Table B. Use this equation:

c_metal = −(c_water · m_water · ΔT_water) / (m_metal · ΔT_metal)

Notes and steps
- c_water = 4.18 J/g·°C.
- Use masses in grams and temperature changes in °C.
- ΔT_water = T_final_water − T_initial_water (usually positive; water warms).
- ΔT_metal = T_final_metal − T_initial_metal (usually negative; metal cools). The leading minus sign in the formula produces a positive c_metal.
- In practice you can use absolute values of the temperature changes and drop the leading minus: c_metal = (4.18 · m_water · |ΔT_water|) / (m_metal · |ΔT_metal|).

What I need from you
For each metal (aluminum wire, steel wire, lead pellets) give:
- m_water (g)
- ΔT_water (°C) or initial and final water temperatures
- m_metal (g)
- ΔT_metal (°C) or initial and final metal temperatures

If you paste those values I will compute c for each metal and give the results (with units J/g·°C). For quick reference, typical literature values are about:
- Aluminum ≈ 0.90 J/g·°C
- Steel (iron-based) ≈ 0.45–0.50 J/g·°C
- Lead ≈ 0.13 J/g·°C

Paste your Table B numbers and I’ll do the calculations.