E = m Lf E = m c θ E = m Lv

E = m c θ E = m c θ E = m Lf E = m c θ

Energy = m x Lf Energy = 10 x 340 Energy = 3,400 J Energy = m x Lf 1.a Energy = 10 x 340 Energy = 3,400 J Energy = m x Lf 1.b Energy = 20 x 340 Energy = 6,800 J

Energy = m x Lf Energy = 10 x 340 Energy = 3,400 J Energy = m x Lf 1.a Energy = 10 x 340 Energy = 3,400 J Energy = m x Lf 1.b Energy = 20 x 340 Energy = 6,800 J

Energy to melt ice = m x Lf 2.a = 5 x 340 = 1700 J Energy to warm up water = m x c x θ = 5 x 4.2 x 50 = 1050 J Total energy required = 1700 J + 1050 J = 2750 J

Energy to melt ice = m x Lf 2.a = 5 x 340 = 1700 J Energy to warm up water = m x c x θ = 5 x 4.2 x 50 = 1050 J Total energy required = 1700 J + 1050 J = 2750 J

Energy to cool water = m x c x θ 2.b = 200 x 4.2 x 20 = 16,800 J Energy change each minute = 16,800 J / 10 min = 1,680 J / Min

Energy to cool water = m x c x θ 2.b = 200 x 4.2 x 20 = 16,800 J Energy change each minute = 16,800 J / 10 min = 1,680 J / Min

Energy to melt ice = m x Lf 3. = 100 x 340 = 34,000 J Energy transferred = power x time = 50 W x t 34,000 J = 680 seconds 34,000 J 50 W

Energy to melt ice = m x Lf 3. = 100 x 340 = 34,000 J Energy transferred = power x time = 50 W x t 34,000 J = 680 seconds 34,000 J 50 W

Energy from cooling rivets = m x c x θ 4a. O C because the block of ice is large and will not all melt! Energy from cooling rivets = m x c x θ 4.b = 170 x 0.9 x 100 = 15,300 J Energy given to melt ice = m x Lf 4.b = m x 340 15,300 J = m = 45 g 15,300 J 340

Energy from cooling rivets = m x c x θ 4a. O C because the block of ice is large and will not all melt! Energy from cooling rivets = m x c x θ 4.b = 170 x 0.9 x 100 = 15,300 J Energy given to melt ice = m x Lf 4.b = m x 340 15,300 J = m = 45 g 15,300 J 340

Energy to change 4g water to steam = m x Lv = 4 x 2,300 = 9,200 J 5.b Energy to change 10 g steam to water = m x Lv Energy to change 10 g steam to water = 10 x 2300 Energy to change 10 g steam to water = 23,000 J Energy given out as water cools to 50 C = m x c x θ = 10 x 4.2 x 50 Total heat given out = 23,000 J + 2,100 J = 25,100 J = 2,100 J

Energy to change 4g water to steam = m x Lv = 4 x 2,300 = 9,200 J 5.b Energy to change 10 g steam to water = m x Lv Energy to change 10 g steam to water = 10 x 2300 Energy to change 10 g steam to water = 23,000 J Energy given out as water cools to 50 C = m x c x θ = 10 x 4.2 x 50 Total heat given out = 23,000 J + 2,100 J = 25,100 J = 2,100 J

Energy transferred = power x time 6. = 3,000 W x 120 s = 360,000 J Energy to change water to steam = m x Lv 360,000 J = m x 2,300 360,000 J = m = 157 g 2,300

Energy transferred = power x time 6. = 3,000 W x 120 s = 360,000 J Energy to change water to steam = m x Lv 360,000 J = m x 2,300 360,000 J = m = 157 g 2,300

Energy from cooling metal = m x c x θ Energy from cooling metal = 200 x 0.5 x 500 Energy from cooling metal = 50,000 J Energy given to vaporize liquid = m x Lf 50,000 = 20 x Lf 50,000 = Lf = 2,500 J/g 20

Energy from cooling metal = m x c x θ Energy from cooling metal = 200 x 0.5 x 500 Energy from cooling metal = 50,000 J Energy given to vaporize liquid = m x Lf 50,000 = 20 x Lf 50,000 = Lf = 2,500 J/g 20