A 50 g sample of an unknown metal is heated to 90.0C. It is placed in a perfectly insulated container along with 100 g of water at an initial temperature of 20C. After a short time, the temperature of both the metal and water become equal at 25C. The specific heat of water is 4.18 J/gC in this temperature range. What is the specific heat capacity of the metal? Record your answer with two significant figures.
5.98
J/gC

0.64 J/g°C

Explanation:

Using the formula;

Q = m × c × ∆T

Where;

Q = amount of heat

m = mass (g)

c = specific heat capacity

∆T = change in temperature (°C)

In this case:

Q (water) = - Q (metal)

mc∆T (water) = - mc∆T (metal)

According to the information in this question,

For water; m = 100g, c = 4.18J/g°C, ∆T = (25°C - 20°C)

For metal; m = 50g, c =?, ∆T = (25°C - 90°C)

mc∆T (water) = - mc∆T (metal)

100 × 4.18 × (25°C - 20°C) = - {50 × c × (25°C - 90°C)}

100 × 4.18 × 5 = - {50 × c × -65}

2090 = -{-3250c}

2090 = 3250c

c = 2090/3250

c = 0.643

c = 0.64J/g°C

they are non-matter

explanation:

they are non-matter. basically, any type of energy or any abstract concept is an example of something that does not matter. other examples include dreams, magnetism, radio, and hate

it becomes positively charged by charging it