What mass of steam initially at 130°C is needed to warm 200g of water in a glass container of mass 100g from 20°C to 50°C.?

Haya Detroit, twende na swali la 6.

Ili kutatua tatizo hili, tutatumia kanuni ya uhifadhi wa nishati, ambapo joto lililopotea na mvuke ni sawa na joto lililopatikana na maji na chombo cha kioo.

Step 1: Badilisha wingi wa maji na chombo cha kioo kutoka gramu kwenda kilogramu.

$$m_w = 200 \text{ g} = 0.2 \text{ kg}$$ $$m_g = 100 \text{ g} = 0.1 \text{ kg}$$

Step 2: Kokotoa joto lililopatikana na maji na chombo cha kioo.
Maji na chombo cha kioo hupashwa moto kutoka $20^\circ C$ hadi $50^\circ C$.

$$\Delta T_{w/g} = 50^\circ C - 20^\circ C = 30^\circ C$$

Joto lililopatikana na maji:

$$Q_w = m_w c_w \Delta T_{w/g} = (0.2 \text{ kg}) \times (4200 \text{ J/KgK}) \times (30 \text{ K})$$ $$Q_w = 25200 \text{ J}$$

Joto lililopatikana na chombo cha kioo:

$$Q_g = m_g c_g \Delta T_{w/g} = (0.1 \text{ kg}) \times (840 \text{ J/KgK}) \times (30 \text{ K})$$ $$Q_g = 2520 \text{ J}$$

Jumla ya joto lililopatikana:

$$Q_{iliyopatikana} = Q_w + Q_g = 25200 \text{ J} + 2520 \text{ J}$$ $$Q_{iliyopatikana} = 27720 \text{ J}$$

Step 3: Kokotoa joto lililopotea na mvuke katika hatua tatu.
Acha $m_s$ iwe wingi wa mvuke.