Question

12. Practice problems with other materials: (Look up values for . Also consider ail digits heat capacity or specific heat shown here to be significant.) a. How hot (what temperature?) would 15 g of aluminum at 20°c get if 90 cal of heat were added? b. How much heat would it take to heat 15 g of copper from 19°C to 31°C? c. What is the specific heat of a metal which requires 80 cal to change 50 g from 20°C to 35°c? d. What is the specific heat of a metal which loses 70 cal of heat when temperature of 40 g of it drops 30°C? e. What is the specific heat of a metal if a 50 g sample cools from 100°C to 25°C while causing 300 g of water to warm from 20°C to 25°C?

Answer 1

a) using the equation q = m s (T2- T1)

q = 90 cal m=15g

s (specific heat capacity)=0.902 J/g ⁰c ( taken from literature)

Since 1cal = 4.18 J

Therefore 90 cal = 90×4.18 J=376.2 J

Putting the value in equation ,we get

376.2 J = 15 g ×0.902J/g ⁰c (T2-20 ⁰c)

T2 = 47.80 ⁰C

Temperature of aluminium will be 47.80 ⁰C.

b) using equation q = ms (T2- T1)

S = 0.385 J/g ⁰c ( taken from literature) m= 15 g

T1= 19 ⁰C and. T2 = 31 ⁰C

Putting the value in equation ,we get

q = 15 g × 0.385 J/ g ⁰c × (31-19) ⁰c

q= 69.3 J

Amount of heat taken by copper= 69.3 J

c) q= ms(T2-T2)

q=80 cal= 80×4.18 J= 334.4 J

m= 50 g T2= 35 ℃ T1 =20℃

Putting the value ,we get

334.4 J = 50 g × s× (35℃-20℃)

S= 0.445 J/g ℃

Specific heat capacity of metal= 0.445J/g ℃

d). q = ms (T2- T1)

q = 70 cal = 70×4.18 J =292.6J

m= 40g. T2- T1= 30 ℃

Putting the value ,we get

292.6 J = 40g × s× 30 ℃

S = 0.244 J/g ℃

Specific heat of metal= 0.244 J / g ℃