Question

800g of ice at -15C is mixed with 100 g of water at 40C cal. final temp.

Answer 1

Given the mass of ice (m2) = 800g = 800/1000 kg
temperature (t2) = - 15°C
The specific heat capacity of ice (c2) = 2100 J / Kg °C

Also given the mass of water (m1) = 100 g = 100/1000 kg
temperature (t1) = 40°C
The specific heat capacity of water(c1) = 4200 J/kg °C

Heat energy gained by cold body = Heat energy lost by the hot body
m1 * c1 (t1 - t) = m2 * c2 (t - t2)
100/1000 * 4200 (40 - t) = 800/1000*2100 (t - (-15))
1/10 * 4200 (40 - t) = 8/10 * 2100 (t+15)
420(40 - t) = 8 * 210 (t+15)
16800 - 420t = 1680t + 25200
16800 - 25200 = 1680t + 420t
- 8400 = 2100t
t = - 8400/2100
t = - 4° C