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

Step-by-step 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* 0.118Cal/g.^oC* \Delta T\\\\\Delta T=(1000)/(100* 0.118)=84.7^oC`

Hence, the change in temperature is 84.7°C