← Zurück zur vorhergehenden Übersicht: ZMK_en
- This topic has 1 reply, 1 voice, and was last updated 1 day, 1 hour ago by .
Viewing 2 posts - 1 through 2 (of 2 total)
- Posts
You can mixed 1 kg of a thickness of not more than 10 mm Ice cream and 1 kg seated Ice cream. What ? Temperature has The Mixed, where: All of it Ice cream of a thickness of not more than 10 mm is? It is important Equations:
Q=c·m·ΔT [KJ]; cw = 4.185kJ/(kg·K)
Lf = 338.8 kJ/kg: WPoor quantity Ice cream melting [kJ/kg]
Solution: - Basic idea: This hot Water cooling The Commission from, This cold Water heated The Commission
- Es Apply the following: Reductions: H for Heiss, K for Cold, M for Mixed
- This Ice cream must first. of a thickness of not more than 10 mm will be (Lf is The required Energy) and can and then First of all on The Mixing temperature heated will be
- Q(100℃ → TM) = Q(Eis schmelzen, 0℃ → TM)
- mH · cw · (100℃ – TM) = mK·Lf + mK · cw · (TM – 0℃ )
- Centigrade in Kelvin , please . Recalculation, mH = mK = m (!)
- m·cw·(373K-TM) = m·Lf +m·cw·(TM-273K)
- TM = [cw·(373k+273k)-Lf]/(2·cw)
- TM = [4.185kJ/kgK·(373k+273k)-335.5kJ]/(2·4.185kJ/kgK) = 283.2 K = 10.2℃
The Mixed would have So what do you mean? One Temperature of about. 10 Degree Centigrade.
Viewing 2 posts - 1 through 2 (of 2 total)
- You must be logged in to reply to this topic.