When 45 g of an alloy, at 25oC, are dropped into 100.0 g of water, the alloy absorbs 956 J of heat. If the final temperature of the alloy is 37oC, what is its specific heat?

The answer to the question is

The specific heat capacity of the alloy = 1.77 J / (g·°C)

Explanation:

To solve this, we list out the given variables thus

Mass of alloy = 45 g

Initial temperature of the alloy = 25 °C

Final temperature of the alloy = 37 °C

Heat absorbed by the alloy = 956 J

Thus we have

ΔH = m·c· (T₂ - T₁) where ΔH = heat absorbed by the alloy = 956 J, c = specific heat capacity of the alloy and T₁ = Initial temperature of the alloy = 25 °C, T₂ = Final temperature of the alloy = 37 °C and m = mass of the alloy = 45 g

∴ 956 J = 45 * C * (37 - 25) = 540 g·°C*c or

c = 956 J / (540 g·°C) = 1.77 J / (g·°C)

The specific heat capacity of the alloy is 1.77 J / (g·°C)