Homework 59

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10.44:

Question:

The initial temperature for each substance, with an unknown mass, is $32.0^{\circ }C$ and absorbs 2450 J of heat. Below, each final temperature is recorded with each substance. With what is given, what is the mass of each substance?

A. Silver ( $T_{F}=81.2^{\circ }C$ )

B. Water ( $T_{F}=54.3^{\circ }C$ )

C. Granite ( $T_{F}=68.9^{\circ }C$ )

What we know:

Initial Temperature( $T_{I}=32.0^{\circ }C$ )
Heat absorbed (2450 J)
Final Temperatures (Silver: $81.2^{\circ }C$ , Water: $54.3^{\circ }C$ , Granite: $68.9^{\circ }C$ )
Specific Heats( $C_{s}$ ) (Silver: $0.235\frac{J}{g\cdot^\circ C}$ , Water: $4.18\frac{J}{g\cdot^\circ C}$ , Granite: $0.79\frac{J}{g\cdot^\circ C}$ ).

What it is asking for: The mass of each substance

Strategy:

Find an equation suitable with the information given
Plug in all the information given in the appropriate places in the equation
Solve and get the mass for each substance

Solution:

Heat= m $C_{s}$ $\Delta T$

m=mass $C_{s}$ = specific heat $\Delta T$ =( $T_{F}-T_{I}$ )

A. Silver

$2450 J= m\cdot (0.235\frac{J}{g\cdot ^{\circ }C}) \cdot (81.2^{\circ }C-32.0^{\circ }C)$

$2450 J= m\cdot (0.235\frac{J}{g\cdot ^{\circ }C}) \cdot (49.2^{\circ }C)$

$2450 J= m\cdot (11.562\frac{J}{g})$

$\frac{2450 J}{11.562\frac{J}{g}}= m$

$m=211.9 g$

B. Water

$2450 J= m\cdot (4.18\frac{J}{g\cdot ^{\circ }C}) \cdot (54.3^{\circ }C-32.0^{\circ }C)$

$2450 J= m\cdot (4.18\frac{J}{g\cdot ^{\circ }C}) \cdot (22.3^{\circ }C)$

$2450 J= m\cdot (93.214\frac{J}{g})$

$\frac{2450 J}{93.214\frac{J}{g}}= m$

$m=26.3g$

C. Granite

$2450 J=m\cdot (0.79\frac{J}{g\cdot ^{\circ }C})\cdot (68.9^{\circ }C-32.0^{\circ }C)$

$2450 J=m\cdot (0.79\frac{J}{g\cdot ^{\circ }C})\cdot (36.9^{\circ }C)$

$2450 J=m\cdot (29.151\frac{J}{g})$

$\frac{2450J}{29.151\frac{J}{g}}=m$

$m=84.05g$

11.40:

Question:

We have a 20.0 L cylinder that is filled with 28.6 g of oxygen gas at the temperature of 401 K. What is the pressure that the oxygen gas is putting on the cylinder?

What we know: