If 25.4 g of water initially at 20.0 C absorbs 115 cal heat what is final temp?
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To determine the final temperature of the water, we can use the equation:
Q = mcΔT
Where:
Q is the heat absorbed by the water
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
Given:
Initial mass of water (m) = 25.4 g
Initial temperature (T1) = 20.0 °C
Heat absorbed (Q) = 115 cal
The specific heat capacity of water (c) is approximately 1 cal/g°C.
To find ΔT, we rearrange the equation as:
ΔT = Q / (mc)
Substituting the given values:
ΔT = 115 cal / (25.4 g * 1 cal/g°C)
Calculating:
ΔT = 115 cal / 25.4 g = 4.5276 °C
Now, we can find the final temperature (T2) by adding ΔT to the initial temperature:
T2 = T1 + ΔT
T2 = 20.0 °C + 4.5276 °C
Calculating:
T2 = 24.5276 °C
The final temperature of the water is approximately 24.5 °C after absorbing 115 cal of heat.
To find the final temperature of water after absorbing heat, we need to use the equation:
𝑞 = 𝑚𝐶∆𝑇
Where:
- 𝑞 is the heat absorbed (in calories)
- 𝑚 is the mass of the water (in grams)
- 𝐶 is the specific heat capacity of water (1 cal/g°C)
- ∆𝑇 is the change in temperature (in °C)
You have the values:
- 𝑞 = 115 cal
- 𝑚 = 25.4 g
- 𝐶 = 1 cal/g°C
Now, rearranging the equation to solve for ∆𝑇:
∆𝑇 = 𝑞 / (𝑚𝐶)
∆𝑇 = 115 cal / (25.4 g * 1 cal/g°C)
Simplifying the equation:
∆𝑇 = 115 / 25.4°C
Calculating:
∆𝑇 ≈ 4.5276°C
Now, to find the final temperature, we add the change in temperature (∆𝑇) to the initial temperature.
Final temperature = Initial temperature + ∆𝑇
Final temperature = 20.0°C + 4.5276°C
Final temperature ≈ 24.53°C (rounded to two decimal places)
Therefore, the final temperature of the water is approximately 24.53°C.