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.

A. 84.7°C
B. 37.0°C
C. 1.43°C
D. 1.18°C

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:

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

Hence, the change in temperature is 84.7°C