Respuesta :

EjerciciosFyQ
The specific heat of iron is 0,113 cal/g·ºC. Clearing the mass in the equation below:

[tex]Q = m\cdot c_e\cdot \Delta T\ \to\ m = \frac{Q}{c_e\cdot \Delta T} = \frac{1\ 450\ cal}{0,113\frac{cal}{g\cdot ^\circ C}\cdot 19,7^\circ C} = \bf 651,36\ g[/tex]
ysabayomi

Answer:

The mass of iron needed will be 669.13 grams

Explanation:

The formula to be used here is that of specific heat equation which is

Q = mc Δt

where Q is the heat energy (in calories according to this question) = 1450 cal

m is the mass of the substance being heated (in this case Iron) = unknown

c is the specific heat of the substance (iron) = 0.11 cal/(g. °C)

Δt is the change in temperature in degree celsius = 19.7 °C

From the specific heat equation provided earlier, the mass of Iron needed can be calculated as

m = Q ÷ (c × Δt)

m = 1450 ÷ (0.11 × 19.7)

m = 1450 ÷ 2.167

m = 669.13 grams

The mass of iron needed will be 669.13 grams

NOTE: Kindly consider the units when solving. The reason for the choice of 0.11 cal/(g. °C) as specific heat is because the heat energy was in calories. If the heat energy was in joules, the specific heat of iron would have been different and the unit would have been j/(g. °C).

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