A 2.00-kg metal object requires 5020 J of heat to raise its temperature from 20.0°C to 40.0°C. What is the spe

Question

A 2.00-kg metal object requires 5020 J of heat to raise its temperature from 20.0°C to 40.0°C. What is the specific heat capacity of the metal?

126 J/(kg • °C)

106 KJ/(kg • °C)

126 cal/(kg • °C)

142 kcal(kg • °C)

Answer 1

The units of specific heat capacity are J / kg•°C. So divide the energy input by the product of the mass and change in temperature. The change in temperature is 40 °C - 20 °C = 20 °C.

C = E / (m*ΔT) = (5020 J) / ((2 kg)(20 °C)) = 125.5 J / (kg-K), which would be 126 if you use only three significant figures.

Answer 2

It is 5020/2/20 = 125.5 J/(kg.C)

Answer 3

a million. warmth = m * c * delta-T the place m = mass in grams, c = particular warmth in cal/(gm - degree ok) and delta-T = replace in temperature. Plug interior the numbers. 2. warmth = m * c * delta-T the place m = mass in grams, c = particular warmth in cal/(gm - degree ok) and delta-T = replace in temperature. Plug interior the numbers.

Answer 4

2kg = 5020j of heat

Raise temp from 20 to 40 degrees.

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