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

the assestment will probley apele with cause and effect