Heat & Specific Heat

If a substance has a high specific heat, what does that imply about its temperature change when heat is added?

• The substance will experience a larger temperature change compared to substances with lower specific heat.

• The substance will experience a smaller temperature change compared to substances with lower specific heat when the same amount of heat is added.

• The substance will not change temperature at all when heat is added.

• The substance will change temperature only if it is in a liquid state.

If 200 J of heat is added to a substance with a specific heat of 0.5 J/g°C and a mass of 10 g, what is the change in temperature?

• 20°C

• 30°C

• 40°C

• 50°C

What is the specific heat of a metal if a 30.00 g piece changes from 25.0 °C to 35.0 °C when it absorbs 350.0 J of energy?

• 0.90 J/g°C

• 1.17 J/g°C

• 1.50 J/g°C

• 2.00 J/g°C

What is the temperature change (∆t) when 100 Joules of heat (Q) is added to 20 grams of copper with a specific heat (c) of 0.39 J/g °C?

• 10.50 °C

• 12.82 °C

• 15.00 °C

• 8.75 °C

True or False: The specific heat is the amount of energy needed to raise the temperature one degree Celsius of 1 gram of a substance.

• True

• False

• It varies by substance

• It is the same for all substances

What is the unit of specific heat?

• J/g°C (Joules per gram per degree Celsius)

• cal/g°C (Calories per gram per degree Celsius)

• kJ/kg·K (Kilojoules per kilogram per Kelvin)

• J/kg·°C (Joules per kilogram per degree Celsius)

What is the effect of adding heat to a substance at its melting point?

• The temperature increases continuously until the substance vaporizes.

• The temperature does not change; instead, the substance undergoes a phase change from solid to liquid.

• The substance becomes a gas immediately without any phase change.

• The temperature decreases as heat is added.

What is the relationship between mass, specific heat, and heat absorbed?

• Q = m * c * ∆T

• Q = m + c + ∆T

• Q = m / c * ∆T

• Q = m * c - ∆T

What is the formula to calculate the heat lost by a substance?

• Q = m * c * (T_initial - T_final)

• Q = m * (T_final - T_initial) * c

• Q = c * (T_initial + T_final) * m

• Q = m * c * T_initial * T_final

Calculate the heat absorbed by 105.0 grams of water when heated from 20.0°C to 99.0°C. The specific heat of water is 4.18 J/g•°C.

• 34673 J

• 25000 J

• 40000 J

• 30000 J