A 250 gram sample of metal is warmed from 50°C to 90°C when 100 calories of heat are added. What is he specific heat capacity of the metal?
100cal=250g*c*40C
c=100/(250*40)= you do it, units are cal/gC
To find the specific heat capacity of the metal, we need to use the formula:
Q = m * c * ΔT
where:
Q = heat energy added (in calories)
m = mass of the metal (in grams)
c = specific heat capacity of the metal (in calories/gram °C)
ΔT = change in temperature (in °C)
Given:
Q = 100 calories
m = 250 grams
ΔT = 90°C - 50°C = 40°C
Now we can rearrange the formula to solve for c:
c = Q / (m * ΔT)
Substituting the given values:
c = 100 calories / (250 grams * 40°C)
First, let's convert 100 calories to joules. 1 calorie is equal to 4.184 joules, so:
100 calories = 100 * 4.184 joules
We get:
c ≈ (100 * 4.184 joules) / (250 grams * 40°C)
To simplify the units, let's convert grams to kilograms:
m = 250 grams = 250 * 0.001 kilograms
Now we have:
c ≈ (100 * 4.184 joules) / (250 * 0.001 kilograms * 40°C)
Simplifying further:
c ≈ 41.84 joules / (0.025 kilograms * 40°C)
Dividing:
c ≈ 41.84 joules / (1 kilogram * 40°C)
c ≈ 41.84 joules / 40 joules
c ≈ 1.046
Therefore, the specific heat capacity of the metal is approximately 1.046 calories/gram °C.