Tuesday 17 Nov

Tuesday 17 Nov:

Target: Understanding Momentum

Lab: Specific Heat

L# 22: 1- 5, 7- 9, 12, 14 - 20 Due Wed 17 Nov

(Calculus: Derivative & Integration )

L# 23: 1-4, 8, 13, 15, 16, 18, 20 Due Fri 19 Nov

(Center of Mass)

Review Monday 22 Nov

Test L# 19 - 23 Tues 23 Nov

Simple Machine Project:

Requirements:

1. Individual

2. Build and its due Wed 24 Nov

(must be made and not just a store example)

2. Any size & any of the 6 types you want.

(as long as you can carry it by your self through the doorway)

3. One page paper explaining your simple machine and its use in History

. Bridge Contest:

Due by Friday Dec 17

Tower Contest:

Due by Friday Jan 7th

Dry Lab:

A) A 10g object is placed into 100⁰C boiling water and brought to equilibrium with it. The object is then placed into 50g of tap temperature water (20⁰C) causing the tap water and the hot object to come to a equilibrium of 25⁰C. What is the specific heat (c_Obj) of the object?

c_water= cal / 1g ⁰c

Q_water = m_water Dt_water c_water

= (50g)(5⁰C)(1g/cal⁰) = 250Cal

Q_obj = m_objDt_objc_obj

250Cal = (10g)(75⁰C)(c_obj)

250Cal / 750g⁰C = 0.33 Cal/g⁰C

B) A 25g object is placed into 100⁰C boiling water and brought to equilibrium with it. The object is then placed into 60g of tap temperature water (20⁰C) causing the tap water and the hot object to come to a equilibrium of 35⁰C. What is the specific heat (c_Obj) of the object?

Q_water = m_water Dt_water c_water

= (60g)(15⁰C)(1g/cal⁰) = 900Cal

Q_obj = m_objDt_objc_obj

900Cal = (25g)(65⁰C)(c_obj)

900Cal / 1,625 = 0.55 Cal/g⁰C

C) A 15g object is placed into 100⁰C boiling water and brought to equilibrium with it. The object is then placed into 40g of tap temperature water (20⁰C) causing the tap water and the hot object to come to a equilibrium of 22⁰C. What is the specific heat (c_Obj) of the object?

Q_water = m_water Dt_water c_water

= (40g)(2⁰C)(1g/cal⁰) = 80Cal

Q_obj = m_objDt_objc_obj

80Cal = (15g)(78⁰C)(c_obj)

80Cal / 1170g⁰C = 0.068 Cal/g⁰C