calories given off when 85g of water cools from 48C to 23C.
Express your answer using two significant figures.
I know the answer is 2100 cal, but I just need to know why the answer is a positive value instead of negative, when it is supposed to be cooling down.
Cooling water means the heat is given off and heat emitted is negative, always. Look at it the other way. If heat were absorbed, it would be positive and the temperature of the water would increase. Right?
I have another question about this problem. I know that dT is (T(final)-T(initial)).
is the T(final)=23C or is it 48C?
since it's from "48C to 23C", shouldn't the final be 23C? and the initial 48C?
To determine the amount of calories given off when water cools from 48°C to 23°C, you can use the specific heat capacity of water. The specific heat capacity of water is approximately 1 calorie per gram per degree Celsius (1 cal/g°C).
First, find the temperature change:
ΔT = T final - T initial = 23°C - 48°C = -25°C
Since the temperature change is negative, which means it is cooling down, it is important to note that the value of ΔT is negative.
Next, calculate the amount of heat energy released:
q = m * c * ΔT
where:
q = heat energy released (calories)
m = mass of water (grams)
c = specific heat capacity of water (cal/g°C)
ΔT = temperature change (°C)
Given:
m = 85g
c = 1 cal/g°C
ΔT = -25°C
Substitute the values into the equation:
q = 85g * 1 cal/g°C * (-25°C)
The negative sign indicates that the heat energy is released or given off during cooling. The negative ΔT value simply signifies the direction of the temperature change, not the sign of the heat energy.
Calculating:
q = -2125 cal
Since we are asked to express the answer using two significant figures, the final answer is 2100 cal.
Therefore, we can conclude that when 85g of water cools from 48°C to 23°C, it releases 2100 calories of heat energy.