Unit 1 - Energy

2025-02-04

q = m c Δ T q = 80.0 (0.90) \times 34 = 2448 g / J

… mercury shc 0.14g/J

m = \frac{Q}{c Δ T} =

375g -15C → gas @ 125C

125—15
$Δ T = 140$
$m c Δ T$

M_{h_{2} o} = 18.02 m (h_{2} o) = 20.8102108768 m o l - 15 *

11.25 1 + 40.6801 5_{2} + 156. 9_{3} + 274.69237 5_{4} + 18.937 5_{5}

38g sample of ice at -11c is placed into 214g of water at 56c

find final temp
water: $m c Δ T$
ice: $h e a t u p - m e lt - h e a t u p$

where $1$ is 56C water and $2$ is the warmed ice

m c Δ T_{1} = m c Δ T_{2} + (m * L) + m c Δ T_{i ce}

(214) (4.184) 56 - (T_{f}) = (38) 2.108 (11) + 333 (38) + 4.184 (38) (T_{f} - 0 c) 50141.056 - T_{f} = 881.144 + 12654 + 158.99200 * T_{f} 50141.056 - T_{f} = 13535.144 + 158.99200 * T_{f}

34.7c