Question

Part C: Energy in Changes of State: CI. Mass of empty calorimeter cup with lid 6-835 107.15 C2. Mass of calorimeter with lid + water C3. Final water temperature Initial water temperature C4. Mass of calorimeter with lid + water + melted ice 135.648 Calculations: CS. Mass of water C6. Temperature change (AT) C7. Calories lost by water Calories used to melt ice C8. Mass of ice that melted C9. Heat of fusion (calories to melt 1 g of ice) Show calculations for C7 Calories lost by water, C8 Mass of ice that melted and C9 Heat of fusion:

Answer 1

Simple 1st law of thermodynamics is been used to solve the above problem..

Post ( (s. Energy in changes of State Mass of water = Moss of calonmates with lid + water - Mass of empty calonmer cop coth led - (107.15 - 6-835)g = 100.32g Temperature chage (AT) - Final waaks temperature - Introl oaks water temperaton - (1-228 -1°C 16 7 Calories loss by water = mass of water x speofic heat of water x temperathon difference - 100.32gx I cal/g x (1-22) = -2106-72 cal Here negative sign indicate that there is loss of energy by water. & From conservation of energy prioople, Calories loss by wooder = colonies geen by used to meltice il calories used to meet = 2106-72 calaries Ce Mass of is that melted = Moss of colosmeter with led & wooden + melted in - Moss of colorimeter coth lid t water = (135.64 - 107.15 g = 28.49g Ca Heat of fusion = Calories used to meet the Ta Total mass of ich 2106.42 calories - 73.94 roolesla = 28.49g