What is the final temperature of a 52 gram glass of water at 2.0 C if it absorbs 3.0 J heat?

To find the final temperature of the glass of water, we can use the formula:

Q = mcΔT

Where:
Q = heat absorbed or released
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature

In this case, the mass of the water (m) is 52 grams and the specific heat capacity (c) of water is 4.18 J/g°C.

We are given that the water absorbs 3.0 J of heat (Q) and the initial temperature (T1) is 2.0°C. We need to find the final temperature (T2).

First, we need to calculate the change in temperature (ΔT):

ΔT = Q / (mc)

ΔT = 3.0 J / (52 g * 4.18 J/g°C)

ΔT = 0.0145°C

Next, we can find the final temperature (T2) by adding the change in temperature to the initial temperature:

T2 = T1 + ΔT

T2 = 2.0°C + 0.0145°C

T2 ≈ 2.0145°C

Therefore, the final temperature of the glass of water is approximately 2.0145°C.

To find the final temperature of the water, we can use the formula:

q = m * c * ΔT

Where:
q = heat absorbed (in Joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)

In this case, we are given:
q = 3.0 J
m = 52 grams
c (for water) = 4.18 J/g°C (specific heat capacity of water)
ΔT = final temperature - initial temperature

We can rearrange the formula to solve for the final temperature:

ΔT = q / (m * c)

Now, substitute the given values into the equation:

ΔT = 3.0 J / (52 g * 4.18 J/g°C)
ΔT = 0.0144 °C/g

Since the ΔT represents the final temperature minus the initial temperature, we can rearrange the equation to solve for the final temperature:

Final temperature = ΔT + initial temperature

We are given the initial temperature as 2.0°C, so substituting the values:

Final temperature = 0.0144 °C/g + 2.0 °C
Final temperature = 2.0144 °C

Therefore, the final temperature of the 52-gram glass of water would be approximately 2.0144°C.

q = mass x specific heat x (Tfinal-Tinitial).