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A 2.40 kg block of ice is heated with 5820 J of heat. The specific heat of water is 4.18 J•g^-1•C^-1. By how much will it’s temperature rise, assuming it does not melt?

2 Answers

2 votes

Final answer:

The temperature of the block of ice will increase by approximately 0.608 °C.

Step-by-step explanation:

To find the temperature change in the block of ice, we need to divide the amount of heat transferred (5820 J) by the mass of the ice (2.40 kg) and the specific heat of water (4.18 J•g-1•C-1). First, convert the mass of the ice to grams: 2.40 kg = 2400 g. Then, use the formula:

ΔT = Q / (m * c)

Where ΔT is the change in temperature, Q is the amount of heat transferred, m is the mass of the substance, and c is the specific heat capacity. Substituting the known values:

ΔT = 5820 J / (2400 g * 4.18 J•g-1•C-1)

Calculating the expression gives a temperature change of ΔT ≈ 0.608 °C.

User Mcstrother
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3 votes

Answer: 0.580 C

Explanation: On Ck-12 I got it right sooo...

User Enashnash
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4.8k points