Question

When a 360 g piece of ole288 of cold water in a is placed in 42528831% the temperature of the alloy decreases by 205°C, while the temperature of the 288water increases by 18.7 °C. Specific heat of water alloy? What is the specific heat (J/g.°C)

- (J/g) 4.184

Answer 1

[AI-generated answer — Bot GPT 3.5]

To find the specific heat of the alloy, we can use the formula:

q = m * c * ΔT

where q is the heat gained or lost, m is the mass, c is the specific heat, and ΔT is the change in temperature.

Since the alloy is losing heat and the water is gaining heat, the equation becomes:

m1 * c1 * ΔT1 = m2 * c2 * ΔT2

Let's label the variables:
m1 = mass of the alloy in grams = 360 g
c1 = specific heat of the alloy (to be determined)
ΔT1 = change in temperature of the alloy = -205 °C (negative since it is losing heat)
m2 = mass of the water in grams = 288 g
c2 = specific heat of water = 4.184 J/g°C
ΔT2 = change in temperature of the water = 18.7 °C

Now we can plug in the values and solve for c1:

360 g * c1 * (-205 °C) = 288 g * 4.184 J/g°C * 18.7 °C

360 * c1 * (-205) = 288 * 4.184 * 18.7

-73860 * c1 = 21919.0528

c1 = 21919.0528 / -73860

c1 ≈ -0.2969 J/g°C

Since specific heat cannot be negative, we'll take the positive value:

c1 ≈ 0.2969 J/g°C

Therefore, the specific heat of the alloy is approximately 0.2969 J/g°C.

Answer 2

[AI-generated answer — Step-by-Step Bot]

To find the specific heat of the alloy, we can use the formula:

q = m * c * ΔT

Where:
q = heat gained or lost by the substance
m = mass of the substance
c = specific heat of the substance
ΔT = change in temperature

We have the following information:

Mass of the alloy (m) = 360 g
Change in temperature of the alloy (ΔT) = -205 °C
Change in temperature of water (ΔT) = 18.7 °C
Specific heat of water (c) = 4.184 J/g °C

Now we can calculate the specific heat of the alloy:

q1 = m * c * ΔT1
q2 = m * c2 * ΔT2

Since the heat lost by the alloy is gained by water in this case (q1 = -q2), we can equate the two equations:

m * c1 * ΔT1 = -m * c2 * ΔT2

Now substitute the given values:

360 g * c1 * -205 °C = -360 g * 4.184 J/g °C * 18.7 °C

Simplify:

c1 * -205 = -4.184 * 18.7

c1 = (-4.184 * 18.7) / -205

c1 ≈ 0.378 J/g °C

Therefore, the specific heat of the alloy is approximately 0.378 J/g °C.