Question

How much heat is required to warm 1.50 kg of sand from 30.0 ∘c to 100.0∘c? the heat capacity of sand is 0.84 j⋅g−1⋅k−1?

Answer 1

8.8 × 10⁴ J

Step-by-step explanation:

Step 1: Given data

mass of sand (m): 1.50 kg × (10³ g/ 1 kg) = 1.50 × 10³ g

Initial temperature: 30.0°C + 273.15 = 303.2 K

Final temperature: 100.0°C + 273.15 = 373.2 K

Change in temperature (ΔT): 373.2 K - 303.2 K = 70.0 K

Heat capacity of sand (c): 0.84 J/g.K

Step 2: Calculate the heat (Q)

We can calculate the required heat using the following expression.

Q = c × m × ΔT

Q = (0.84 J/g.K) × (1.50 × 10³ g) × 70.0 K = 8.8 × 10⁴ J

Answer 2

Answer is: the amount of heat required to warm the sand is 88.2 kJ.
m(sand) = 1,5 kg · 1000 g/kg = 1500 g.
ΔT = 100°C - 30°C = 70°C.
C(sand) = 0,84 J/g·°C.
Q = m(sand) · ΔT · C(sand).
Q = 1500 g · 70°C · 0,84 J/g·°C.
Q = 88200 J ÷ 1000 J/kJ= 88,200 kJ.