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.
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
1 answer