Question

How much heat, in oules and in, calories, is required to heat a 28.4-g ( 1-oz) ice cube from -23.0 to -1.0 degree celcius?

Answer 1

Answer: The heat required by iron is 1307.11 Joules or 312.4 Cal.

Step-by-step explanation:

To calculate the amount of heat released or absorbed, we use the equation:

`q=mc\Delta T`

where,

q = heat absorbed

m = mass of ice = 28.4 g

c = specific heat capacity of ice = 0.5Cal/g.°C

`\Delta T` = change in temperature =
`T_2-T_1=-1^oC-(-23^oC)=22^oC`

Putting values in above equation, we get:

`q=28.4g* 0.5Cal/g.^oC* (22^oC)\\\\q=312.4Cal`

Converting the value from calories to joules, we use the conversion factor:

1 J = 0.239 Cal

So,
`312.4Cal=(1J)/(0.239Cal)* 312.4Cal=1307.11J`

Hence, the heat required by ice is 1307.11 Joules or 312.4 Cal.