If you me ill give you mores. plzz .at 1 atm, how much energy is required to heat 37.0 g of h2o(s) at –10.0 °c to h2o(g) at 155.0 °c? constants can be found here.

116.3 kJ

Step-by-step explanation:

Three heat transfers are involved

q = Heat to warm ice + heat to melt ice + heat to warm water + heat to evaporate water + heat to warm steam

q =      q₁      +     q₂        +     q₃       +     q₄          +     q₅

q = mC₁ΔT₁ + mΔH_fus + mC₃ΔT₃ + mΔH_vap + mC₅ΔT₅

Step 1: Calculate q₁

m = 37.0 g

C₁ = 2.010 J·°C⁻¹g⁻¹

ΔT₁ = T_f – T_i  

ΔT₁ = 0.0 – (-10.0)

ΔT₁ = 10.0 °C  

q₁ = 37.0 × 2.010 × 10.0  

q₁= 743.7 J

q₁= 0.7437 kJ

Step 2. Calculate q₂

ΔH_fus = 334 J/g

q₂ = 37.0 × 334

q₂ = 12 360 J

q₂ = 12.36 kJ

Step 3: Calculate q₃

C₃ = 4.179 J·°C⁻¹g⁻¹

ΔT₃ = T_f – T_i  

ΔT₃ = 100 – 0  

ΔT₃ = 100 °C

q₃ = 37.0 × 4.179 × 100  

q₃ = 15 460 J

q₃ = 15.46 kJ

Step 4. Calculate q₄

ΔH_vap = 2260 J/g

q₄ = 37.0 × 2260

q₄ = 83 620 J

q₄ = 83.62 kJ

Step 5. Calculate q₅

C¬₅ = 2.010 J·°C⁻¹g⁻¹

ΔT₅ = T_f – T_i  

ΔT₅ = 155.0 – 1000  

ΔT₅ = 55.0 °C

q₅ = 37.0 × 2.010 × 55

q₅ = 4090 J

q₅ = 4.090 kJ

Step 6. Calculate q

q = 0.7437 + 12.36 + 15.46 + 83.62 + 4.090

q = 116.3 kJ

The heat required is 116.3 kJ.