The specific heat for water (c) is 4186 J/(kg * C°).
the equation Q = (m)(c)(Δ T)
For Trail 1:
Hot water mass = 0.25 kg
Cold water mass = 0.25 kg
Use the equation Q = (m)(c)(Δ T)to calculate the heat gained by the cold water for each tria 1 Q1 = (m)(c)(Δ T)
Q1 = (0.25 kg )(4186)(36 C°)
Q1 = 37674 J -------------------(1)
Use the equation Q = (m)(c)(Δ T)to calculate the heat "lost" by the hot water for each trial 1 Q2 = (m)(c)(Δ T)
Q2 = (0.25 kg )(4186 J/(kg * C°))(-42 C°)
Q2 = - 43953 J -------------------(2)
the total heat given off by warmer substances equals the total heat energy gained by cooler substances = Q1 - Q2
= 37674 - (-43953) J
=81,627 J
For Trail 2:
Hot water mass = 0.40 kg
Cold water mass = 0.20 kg
Use the equation Q = (m)(c)(Δ T)to calculate the heat gained by the cold water for each tria 2 Q1 = (m)(c)(Δ T)
Q1 = (0.40 kg )(4186)(36 C°)
Q1 = 60,278.4 J -----------------(3)
Use the equation Q = (m)(c)(Δ T)to calculate the heat "lost" by the hot water for each trial 2 Q2 = (m)(c)(Δ T)
Q2 = (0.20 kg )(4186 J/(kg * C°))(-42 C°)
Q2 = - 35,162.4 J ------------------(4)
the total heat given off by warmer substances equals the total heat energy gained by cooler substances = Q1 - Q2
=95,440.8 J
For Trail 3:
Hot water mass = 0.15 kg
Cold water mass = 0.30 kg
Use the equation Q = (m)(c)(Δ T)to calculate the heat gained by the cold water for each tria 3 Q1 = (m)(c)(Δ T)
Q1 = (0.15 kg )(4186)(36 C°)
Q1 = 22,604.4 J -------------------(5)
Use the equation Q = (m)(c)(Δ T)to calculate the heat "lost" by the hot water for each trial 3 Q2 = (m)(c)(Δ T)
Q2 = (0.30 kg )(4186 J/(kg * C°))(-42 C°)
Q2 = - 52,743.6 J --------------------(6)
the total heat given off by warmer substances equals the total heat energy gained by cooler substances = Q1 - Q2
=75,348 J