A 0.25 kg cube of iron is heated and its temperature rises from 23.0 C to 45.0 C. How much heat entered the cube in Joules and in Calories?
Q = .25 * specific heat of iron * (45-23)
You have those specific heats in Joules and Calories in your text
just multiply then??
Yes
iron
470 J/kg K
iron
470 J/kg K
108 cal/kg K
To find the amount of heat (Q) that entered the cube, we can use the formula:
Q = mcΔT
where m is the mass of the cube, c is the specific heat capacity of iron, and ΔT is the change in temperature.
First, let's calculate the change in temperature:
ΔT = Tfinal - Tinitial
= 45.0°C - 23.0°C
= 22.0°C
Next, we need to look up the specific heat capacity of iron. The specific heat capacity represents the amount of heat energy needed to raise the temperature of 1 kg of substance by 1 degree Celsius. The specific heat capacity of iron is 0.449 J/g°C or 0.107 cal/g°C.
Now, let's calculate the heat (Q) in Joules:
Q = mcΔT
= (0.25 kg) x (0.449 J/g°C) x (22.0°C)
= 2.47 J
Therefore, the amount of heat that entered the cube is 2.47 Joules.
To convert this value to calories, we can use the conversion factor: 1 calorie = 4.184 J.
Now, let's calculate the heat (Q) in Calories:
Q (calories) = Q (Joules) / 4.184 J
Q (calories) = 2.47 J / 4.184 J
= 0.59 Calories
Therefore, the amount of heat that entered the cube is approximately 0.59 Calories.