500 J of heat are added to 59 g of water initially at 20°C.
(a) How much energy is this in calories?
(b) What is the final temperature of the water?
I know that to find (a) I have to use the constant 4.19 J and divide it into the 500 J yielding 119.33 calories. How do I find final temperature?
1J =0.2388 cal
500 J = 119.4229 cal.
Q =mcΔT,
ΔT = Q/mc = 500/0.059•4180 =02.03 oC
T1 = 20+ 2.03 = 22.03 oC
To find the final temperature of the water, you can use the principle of heat transfer, specifically the formula for heat transfer:
q = mcΔT
Where:
q is the heat transferred (in joules or calories),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/g°C or cal/g°C), and
ΔT is the change in temperature (in °C).
To solve for the final temperature, rearrange the formula:
ΔT = q / (mc)
Given:
q = 500 J
m = 59 g
c = specific heat capacity of water (approximately 4.18 J/g°C)
Substituting the known values:
ΔT = 500 J / (59 g * 4.18 J/g°C)
Performing the calculation:
ΔT ≈ 2.01 °C
The change in temperature is approximately 2.01 °C. To find the final temperature, add this change to the initial temperature of the water (20°C):
Final temperature = Initial temperature + ΔT
Final temperature = 20°C + 2.01°C
Therefore, the final temperature of the water is approximately 22.01°C.