Question

The specific heats and densities of several materials are given below: Material Specific Heat (cal/g·°C) Density (g/cm3) Brick 0.220 2.0 Concrete 0.270 2.7 Steel 0.118 7 Water 1.00 1.00 Calculate the change in temperature produced by the addition of 1 kcal of heat to 100 g of steel.

Answer 1

Answer: The change in temperature is 84.7°C

Explanation:

To calculate the change in temperature, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed = 1 kCal = 1000 Cal    (Conversion factor: 1 kCal = 1000 Cal)

m = mass of steel = 100 g

c = specific heat capacity of steel = 0.118 Cal/g.°C

$\Delta T$ = change in temperature = ?

Putting values in above equation, we get:

$1000cal=100g\times 0.118cal/g^oC\times \Delta T\\\\\Delta T=\frac{1000cal}{100g\times 0.118cal/g^oC}\\\\\Delta T=84.7^oC$

Hence, the change in temperature is 84.7°C

Answer 2

The change in the temperature produced by the addition of 1 KCal of heat to 100 g of steel is 84.7 °C

From the question given above, the following data were obtained:

Mass of steel (M) = 100 g

Specific heat capacity of steel (C) = 0.118 Cal/gºC

Heat (Q) = 1 KCal = 1000 Cal

Change in temperature (ΔT) =?

The change in the temperature of the steel can be obtained as follow:

Q = MCΔT

1000 = 100 × 0.118 × ΔT

1000 = 11.8 × ΔT

Divide both side by 11.8

ΔT = 1000 / 11.8

ΔT = 84.7 °C

Therefore, the change in the temperature of the steel is 84.7 °C

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