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Ksivusya [100]
4 years ago
10

A platinum ball weighing 100 g is removed from a furnace and dropped into 400 g of water at 0 degree C. If the equilibrium tempe

rature is 10 degrees C, and the specific heat of platinum is 0.04 cal/g degree C, what must have been the temperature of the furnace? Neglect the effect of the mass of the calorimeter.
Physics
1 answer:
kakasveta [241]4 years ago
3 0

Answer:

T = 1010 degree Celsius

Explanation:

mass of ball (Mb) = 100 g

mass of water (Mw) = 400 g

temp of water = 0 degree

specific heat of platinum (C) = 0.04 cal/g degree celsius

we can calculate the temperature of the furnace from the equation before

Mb x C x (temp of furnace (T) - equilibrium temp) = Mw x (equilibrium temp - temp of furnace)

100 x 0.04 x ( T - 10) = 400 x (10 - 0)

4 (T - 10) = 4000

T - 10 = 1000

T = 1010 degree Celsius

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. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of
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Answer:

See Explanation

Explanation:

Given

s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}

Solving (a): Units and dimension of s_0

From the question, we understand that:

s \to L --- length

t \to T --- time

Remove the other terms of the equation, we have:

s=s_0

Rewrite as:

s_0=s

This implies that s_0 has the same unit and dimension as s

Hence:

s_0 \to L --- dimension

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Solving (b): Units and dimension of v_0

Remove the other terms of the equation, we have:

s=v_0t

Rewrite as:

v_0t = s

Make v_0 the subject

v_0 = \frac{s}{t}

Replace s and t with their units

v_0 = \frac{L}{T}

v_0 = LT^{-1}

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v_0 \to LT^{-1} --- dimension

v_0 \to m/s --- unit

Solving (c): Units and dimension of a_0

Remove the other terms of the equation, we have:

s=\frac{a_0t^2}{2}

Rewrite as:

\frac{a_0t^2}{2} = s_0

Make a_0 the subject

a_0 = \frac{2s_0}{t^2}

Replace s and t with their units [ignore all constants]

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a_0 = LT^{-2

Hence:

a_0 = LT^{-2 --- dimension

a_0 \to m/s^2 --- acceleration

Solving (d): Units and dimension of j_0

Remove the other terms of the equation, we have:

s=\frac{j_0t^3}{6}

Rewrite as:

\frac{j_0t^3}{6} = s

Make j_0 the subject

j_0 = \frac{6s}{t^3}

Replace s and t with their units [Ignore all constants]

j_0 = \frac{L}{T^3}

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Ignore other terms of the equation, we have:

s=\frac{ct^5}{120}

Rewrite as:

\frac{ct^5}{120} = s

Make c the subject

c = \frac{120s}{t^5}

Replace s and t with their units [Ignore all constants]

c = \frac{L}{T^5}

c = LT^{-5}

Hence:

c \to LT^{-5} --- dimension

c \to m/s^5 --- units

4 0
3 years ago
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